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ELEMENTS 

OF 

PLANE   TRIGONOMETRY 


By  D.  A,  MURRAY,  Ph.D, 

Professor  of  Applied  Mathematics  in  McGill  University. 


INTRODUCTORY  COURSE  IN  DIFFERENTIAL  EQUA- 
TIONS, for  Students  in  Classical  and  Engineer- 
ing Colleges.    Pp.  xvi  +  236. 

A    FIRST    COURSE    IN    INFINITESIMAL    CALCULUS. 

Pp.  xvii  +  439. 
DIFFERENTIAL    AND    INTEGRAL    CALCULUS.      Pp. 

xviii  +  491. 

PLANE  TRIGONOMETRY,  for  Colleges  and  Second- 
ary Schools.    With  a  Protractor.    Pp.  xiii  +  212. 

SPHERICAL  TRIGONOMETRY,  for  Colleges  and 
Secondary  Schools.    Pp.  x  + 114. 

PLANE  AND  SPHERICAL  TRIGONOMETRY.  In  One 
Volume.    With  a  Protractor.     Pp.  349. 

PLANE  AND  SPHERICAL  TRIGONOMETRY  AND 
TABLES,     in  One  Volume.     Pp.  448. 

PLANE  TRIGONOMETRY  AND  TABLES.  In  One  Vol- 
ume.   With  a  Protractor.    Pp.  324. 

LOGARITHMIC  AND  TRIGONOMETRIC  TABLES.  Five- 
place  and  Four-place.    Pp.  99. 


NEW  YORK:  LONGMANS,  GREEN,  &  CO, 


ELEMENTS 


OF 


PLANE   TRIGONOMETRY 


BY 


DANIEL  A.    MURRAY,    Ph.D. 

Professor  of  Applied  Mathematics  in  MgGill  University 


LONGMANS,    GREEN,    AND    CO 

FOURTH  AVENUE  &  30TH  STREET,  NEW  YORK 
LONDON,    BOMBAY,    AND    CALCUTTA 

1912 


REPLACING 

? G2 3  o  I 


Copyright,   1911, 

BY 

LONGMANS,    GREEN,    &   CO. 


First  Edition,  July,  1911 

Reprinted,  August,    1912 

October,  1912. 


THE  SCIENTIFIC   PRESS 

HOBERT   DRUMMOND   AND   COMPANV 

BROOKLYN,    N-    Y. 


QAssz 
/In 


PREFACE 


This  text-book  is  shorter  than  my  former  text-book, 
entitled  "  Plane  Trigonometry,"  by  the  omission  of  many 
of  the  notes  and  several  of  the  topics  in  that  book  and  by 
the  more  condensed  treatment  of  other  topics.  There  is 
also  a  marked  difference  in  arrangement.  Thus,  radian 
measure,  the  periodicity  of  the  trigonometric  functions, 
their  general  values,  and  their  graphs,  and  the  inverse  trigo- 
nometric functions,  which  are  discussed  in  the  later  chapters 
of  the  "Plane  Trigonometry/ '  are  treated  in  the  earlier 
chapters  of  the  "Elements  of  Plane  Trigonometry."  The 
line  definitions  of  the  functions  are  explained  more  fully, 
and  the  unit  circle  is  used  to  a  greater  extent,  in  this  book 

than  in  the  former  one. 

D.  A.  Murray. 

May  1,  1911. 


M552649 


CONTENTS 


CHAPTER  I 
Trigonometric  Functions  of  Acute  Angles 

ART.  PAGE 

1.  Angle  denned 1 

2.  Degree  measure 2 

3.  Trigonometric  functions  denned  for  acute  angles 2 

4.  Problems 5 

5.  Trigonometric  functions  of  45°,  60°,  30°,  0°,  90°.     Variation  of 

functions 8 

6.  Relations  between  the  trigonometric  functions  of  an  acute  angle 

and  those  of  its  complement 12 

7.  Relations  between  the  trigonometric  functions  of  an  acute  angle  .  13 

CHAPTER  II 
Solution  of  Right-angled  Triangles 

8.  Solution  of  a  triangle 18 

9.  Cases  in  the  solution  of  right-angled  triangles 19 

10.  Projection  of  a  straight  line  upon  another  straight  line 24 

11.  Measurement  of  heights  and  distances 25 

12.  Solution  of  isosceles  triangles 28 

13.  Related  regular  polygons  and  circles 29 

14.  Problems  requiring  a  knowledge  of  the  points  of  the  mariner's 

compass 32 

15.  Examples  in  the  measurement  of  land 33 

vii 


vili  CONTENTS 

CHAPTER  III 
Angles  in  General  and  Their  Trigonometric  Functions 

ART.  PAGE 

16.  General  definition  of  an  angle 36 

17.  Measurement  of  angles 39 

18.  Value  of  a  radian 40 

18a.  Motion  of  a  particle  in  a  circle. 42 

19.  The  convention  for  signs  of  lines  in  a  plane 43 

20.  Trigonometric  functions  defined  for  angles  in  general 44 

21.  Line   definitions    of    the    trigonometric    functions.     Geometrical 

representation  of  the  functions 48 

22.  Changes  in  the  values  of  the  functions  when  the  angle  varies. 

Limiting  values  of  the  functions 51 

23.  Periodicity  of  the  trigonometric  functions 55 

24.  Graphs  of  the  functions 56 

25.  Relations  between  the  trigonometric  functions  of  an  angle 60 

26.  Functions  of  -A,  90°,  TA,  180,  TA  in  terms  of  functions  of  A .  62 

27.  Reduction  of  trigonometric  functions  of  any  angle  to  functions 

of  acute  angles 66 

CHAPTER  IV 

General  Values.     Inverse  Trigonometric  Functions. 
Trigonometric  Equations 

28.  General  values 68 

29.  Inverse  trigonometric  functions 72 

30.  Trigonometric  equations.     Trigonometric  identities 75 

CHAPTER  V 

Trigonometric  Functions  of  the  Sum  and  Difference  of  Two 

Angles 

general  formulas 

31.  To  deduce  sin  (A  +  B),  cos  (A  +  £) 78 

32.  To  deduce  sin  (A  -B),  cos  (A  -B) 81 

33.  Fundamental  formulas «, 83 

34.  To  deduce  tan  (A+B),  tan  (A-B),  cot  (A  +  £),  cot  (A-B) 84 

35.  To  deduce  sin  2 A,  cos  2 A,  tan  2A 85 

36.  Transformation  formulas 88 


CONTENTS  ix 


CHAPTER  VI 

Relations  between  the  Sides  and  Angles  of  a  Triangle 
ah¥.  page 

37.  Notation.     Simple  geometrical  relations 94 

38.  The  law  of  sines ■ 94 

39.  The  law  of  cosines ' 96 

40.  The  law  of  tangents 97 

41.  Functions  of  the  half -angles  of  a  triangle  in  terms  of  its  sides. ...  98 

CHAPTER  VII 

Solution  of  Oblique  Triangles 

42.  Cases  for  solution.     General  remarks  on  methods  of  solution 101 

43.  Case     I.  Given  one  side  and  two  angles 102 

44.  Case    II.  Given  two  sides  and  an  angle  opposite  to  one  of  them. .  103 

45.  Case  III.  Given  two  sides  and  their  included  angle 107 

46.  Case  IV.  Given  three  sides 108 

47.  Use  of  logarithms  in  the  solution  of  triangles 109 

48.  Cases  I,  II,  logarithms  used 109 

49.  Case  III,  logarithms  used Ill 

J>0.  Case  IV,  logarithms  used 112 

51.  Problems  in  heights  and  distances 113 

CHAPTER  VIII 

Miscellaneous  Theorems 

52.  Area  of  a  triangle 117 

53.  Area  of  a  circular  sector 118 

54.  Circles  connected  with  a  triangle 119 

55.  Relations  between  the  radian  measure,  the  sine,  and  the  tangent 

of  certain  angles 123 

Answers  to  the  Examples ...t 128 


ELEMENTS 


OF 


PLANE  TRIGONOMETRY 


CHAPTER  I 

TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES 

1.  Angle    defined.     An   angle   XOP   is    the    amount    of 
turning  which  a  line  makes  when  it  revolves  about  0  from 


Fig.  1. 


the  position  OX  into  the  position  OP  (Figs.  1,  2).     Accord- 
ingly (Fig.  2), 

one-fourth  of  a  revolution  =  a  right  angle; 
a  complete  revolution        =  four  right  angles. 


2  ELEMENTS  OF  PLANE  TRIGONOMETRY 

When  the  amount  of  turning  is  less  than  one-fourth  a  revo- 
lution (i.e.,  less  than  right  angle  XOY),  the  angle  is  called 
an  acute  angle. 

2.  Degree  (or  sexagesimal)  measure  of  angles.  Since  all 
right  angles  are  equal,  a  right  angle  may  be  chosen  as  the 
unit  of  measurement.  A  right  angle,  however,  is  incon- 
veniently large  as  a  unit.  Accordingly  a  ninetieth  part  of  a 
right  angle,  called  a  degree,  is  taken  for  unit.  Degrees  are 
divided  into  minutes,  and  minutes  into  seconds,  according 
to  the  following  table  of  angular  measure : 

60  seconds  =  1  minute, 
60  minutes  =  1  degree, 
90  degrees  =  1  right  angle. 

Degrees,  minutes,  and  seconds  are  denoted  by  symbols: 
thus,  23  degrees  17  minutes  20  seconds  is  written  23°  17'  20". 

3.  Trigonometric    functions    (defined    for    acute    angles). 

From  any  point  P  in  one  of  the  lines  bounding  an  angle  A 
(Figs.  3,  4,  5)  draw  a  perpendicular  PM  to  the  other  bound- 
ing line.     (The  angles  A  in  Figs.  3,  4,  5  are  equal.) 


In  any  of  these  triangles  AMP  there  can  be  formed  six 
ratios  with  the  lines  AM,  MP,  AP,  viz.: 

MP     AM     MP_     AM     AP_     AP^ 
~AP'    AP'    AM'    MP'    AM'    MP' 


TRIGONOMETRIC  FUNCTIONS    OF  ACUTE  ANGLES 


Since  all  the  triangles  AMP  above  are  similar,  each  of 
these  ratios  has  the  same  value  whatever  be  the  position 
of  P  on  a  bounding  line  of  the  angle.  These  six  ratios  are 
called  trigonometric  functions  of  the  angle  A,  and  are  given 
names  as  follows: 

MP  . 

is  called  the      sine       of  the  angle  A ; 


AP 
AM  . 


-7-p  is  called  the     cosine     of  the  angle  A ; 
-j-jn?  is  called  the   tangent    of  the  angle  A ; 

-rjrp  is  called  the  cotangent  of  the  angle  A ; 

AP 
AM 


is  called  the    secant     of  the  angle  A ; 


AP 


(1) 


-TTp  is  called  the  cosecant  of  the  angle  A. 

Short  symbols  for  these  functions  and  definitions  appli- 
cable for  any  right-angled  triangle  are  given  in  (2) : 

opposite  side 
hypotenuse  ' 


•       A    I      MP\ 

sin  A  {=jp) 


COS 


tan  A 


cot  A 


sec 


/  AM\     adjacent  side 

\  AP/~  hypotenuse  ' 

/  MP\     opposite  side 

\  AM/  ~  adjacent  side' 

/  AM\     adjacent  side 

\  MP)  ~  opposite  side ' 

/  AP\      hypotenuse 

\  AM/  ~  adjacent  side' 


cosed  \  =  mp) 


hypotenuse 
opposite  side* 


(2) 


4  ELEMENTS  OF  PLANE  TRIGONOMETRY 

The  symbol  esc  A  is  also  used  for  cosecant  A. 

From  the  definitions  of  these  functions  and  the  proper- 
ties of  similar  triangles  the  following  properties  are  easily 
deduced : 

(1)  To  each  value  of  an  angle  there  corresponds  but  one 
value  of  each  trigonometric  function. 

(2)  To  each  value  of  a  trigonometric  function  there  corre- 
sponds but  one  value  of  an  acute  angle. 

(3)  Two  unequal  acute  angles  have  different  values  for  each 
trigonometric  function. 

The  values  of  the  trigonometric  functions  for  angles 
from  0°  to  90°  are  arranged  in  tables.  These  values,  which 
may  be  given  to  four,  five,  six  or  seven  places  of  decimals, 
are  called  the  Natural  sines,  Natural  cosines,  etc.  The 
logarithms  of  these  values  of  sines  and  cosines  (with  10  added) 
are  called  Logarithmic  sines,  Logarithmic  cosines,  etc. 

In  addition  to  functions  (1),  (2),  the  following  are  occa- 
sionally used: 

versed  sine  of  A  =     vers  A  =  l  —  cos  A ; 
coversed  sine  of  A  =  covers  A  =  \— sin  A. 


EXAMPLES 

1.  Suppose  that  the  line  OP  (Fig.  2)  revolves  about  0  in 
a  counter-clockwise  direction,  starting  from  the  position  OX; 
show  that,  as  the  angle  XOP  increases,  its  sine,  tangent,  and 
secant  increase,  and  its  cosine,  cotangent,  and  cosecant 
decrease.  Test  this  conclusion  by  an  inspection  of  a  table  of 
Natural  functions. 

2.  Find  by  tables,  sin  17°  40',  sin  76°  43',  cos  18°  10', 
cos  61°  37',  tan  79°  37'  30",  cot  72°  25' 30".  Log  sin  37°  20', 
log  cos  71°  25',  log  tan  79°  30'  20". 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES    5 

3.  Find  the  angles  corresponding  to  the  following  Natural 
and  Logarithmic  functions: 

sine=  .15327,  sine=  .62175, 

cosine  =.85970,  cosine  =.6 1497, 

tangent  =  .42482,       tangent  =  .60980, 

Log  sine =9.79230, 

Log  cosine =9.96611, 

Log  tangent  =  9.82120. 

4.  Problems.     The  student   is   recommended  to  try  to 
solve  Exs.  1,  2,  3,  4  without  help  from  the  book. 


EXAMPLES 

1.  Construct  the  acute  angle  whose  cosine  is  § . 
its    other    trigonometric    functions?     Find 
the  number  of  degrees  in  the  angle. 

The  required  angle  is  equal  to  an  angle 
in  a  right-angled  triangle,  in  which  "the 
side  adjacent  to  the  angle  is  to  the  hypot- 
enuse in  the  ratio  2:3."  Construct  a 
right-angled  triangle  AST  which  has  side 
AS=2,  and  hypotenuse  A T=  3.  The  angle 
A  is  the  angle  required,  for  cos  A  =  § . 


What  are 


Now 


ST=  V32-22=  V5  =  2.2361. 


Fig.  6. 


Hence,  the  other  functions  are 

VE  V5 

sin  A  =  —-=.7454,    tan  A  =  -^-=1.1180, 


cot  A  =  -7=  =.8944, 

v5 


sec  A  =  -=1.5000, 


cosec  A  =  — 7==  1.3416. 

V5 


The  measure  of  the  angle  can  be  found  in  either  one  of  two 
ways,  viz.:  (a)  by  measuring  the  angle  with  the  protractor; 
(b)  by  finding  in  the  table  the  angle  whose  cosine  is  j  or  .6667. 
The  latter  method  shows  that  A  =  48°  11.4'.  [Compare  the 
result  obtained  by  method  (a)  with  the  value  given  by  method 


ELEMENTS  OF   PLANE  TRIGONOMETRY 


a 


2.  A  right-angled  triangle  has  an  angle  whose  cosine  is  §, 
and  the  length  of  the  hypotenuse  is  50  ft. 
Find  the  angles  and  the  lengths  of  the  two 
sides. 

By  method  shown  in  Ex.  1,  construct 
C;an   angle   A   whose    cosine  is  f.     On  one 
boundary  line  of  the  angle  take   a  length 
AG  to  represent  50  ft.      Draw  GK  perpen- 
dicular to  the  other  boundary  line. 
7. 

cos  A  =  l  =.6666.  .  .  , 

.-.     A =48°  11.4', 


.\     J3=  90  -A  =  41°  48.6'. 


AK 


cos  A  =  —  =  .6666.  .  ., 

AG  ' 


.*.     Ai£=50X.6666.  .  ., 


=  33.333 


sin  A  = 


(Ex.  1) 


KG    Vb 
AG~~T' 


V5 
KG=  —  X  50  =  37.27. 

o 


The  problem  may  also  be  solved  graphically  as  follows: 
Measure  angles  A,  G  with  the  protractor.  Measure  AK,  KG 
directly  in  the  figure. 

3.  A  ladder  24  ft.  long  is  leaning  against 
the  side  of  a  building,  and  the  foot  of  the 
ladder  is  distant  8  ft.  from  the  building  in  a 
horizontal  direction.  What  angle  does  the 
ladder  make  with  the  wall?  How  far  is  the 
end  of  the  ladder  from  the  ground? 

Graphical    method.     Let    AC    represent    the 

ladder,  and  BC  the  wall.      Draw   AC,   AB,    to 

scale,  to  represent  24  ft.  and  8  ft.  respectively. 

Measure     angle     ACB     with     the    protractor. 

Fig.  8.         Measure  BC  directly  in  the  figure. 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES  ^ 

Method  of  computation. 
5C=x/IC2-ZB2=V576-64=V512= 22.63  ft. 
smACB=j^=  ^=.33333, 

.-.     ACB=19°28.2'. 

4.  Find  tan  40°  by  construction  and  measurement.     With 
the  protractor  lay  off  an  angle  SAT  equal 
to  40°.     From  any  point  P  in  AT  draw  PR 
perpendicularly  to  AS.     Then  measure  AR, 
RP,  and  substitute  the  values  in  the  ratio, 

-pp 

tan  40°  =  -Tp-     Compare  the  result  thus  ob- 

tained    with    the  value  given   for   tan  40° 
in   the   tables. 

5.  Construct  the  angle  whose  tangent  is  f.  Find  its  other 
functions.  Measure  the  angle  approximately,  and  compare 
the  result  with  that  given  in  the  tables.  Draw  a  number  of 
right-angled,  obtuse-angled,  and  acute-angled  triangles,  each 
of  which  has  an  angle  equal  to  this  angle. 

6.  Similarly  for  the  angle  whose  sine  is  $;  and  for  the 
angle  whose  cotangent  is  3. 

7.  Similarly  for  the  angle  whose  secant  is  2£;  and  for  the 
angle  whose  cosecant  is  3J. 

8.  Find  by  measurement  of  lines  the  approximate  values 
of  the  trigonometric  functions  of  30°,  40°,  45°,  50°,  55°,  60°, 
70°;   compare  the  results  with  the  values  given  in  the  tables. 

//  any  of  the  following  constructions  asked  for  is  impossible, 
explain  why  it  is  so. 

9.  Construct  the  acute  angles  in  the  following  cases:  (a) 
when  the  sines  are  %,  2,  f ;  (6)  when  the  cosines  are  i,  £,  3; 
(c)  when  the  tangents  are  3,  4,  § ;  (d)  when  the  cotangents  are 
4,  |;  (e)  when  the  secants  are  2,  3,  i;  (/)  when  the  cosecants 
are  3,  4, 


8 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


10.  Find  the  other  trigonometric  functions  of  the  angles  in 
Ex.  9.  Find  the  measures  of  these  angles,  (a)  with  the  pro- 
tractor, (b)  by  means  of  the  tables. 

11.  What   are  the   other  trigonometric   functions   of  the 

angles:    (1)  whose  sine  is  — ;    (2)  whose  cosine  is  — ;  (3)  whose 

.    a  .a  a 

tangent  is  r-;    (4)  whose  cotangent  is  =-;  (5)  whose  secant  is  -; 

a 
(6)  whose  cosecant  isr? 

12.  A  ladder  32  ft.  long  is  leaning  against  a  house,  and 
reaches  to  a  point  24  ft.  from  the  ground.  Find  the  angle 
between  the  ladder  and  the  wall. 

13.  A  man  whose  eye  is  5  ft.  8  in.  from  the  ground  is  on 
a  level  with,  and  120  ft.  distant  from  the  foot  of  a  flag  pole 
45  ft.  8  in.  high.  What  angle  does  the  direction  of  his  gaze, 
when  he  is  looking  at  the  top  of  the  pole,  make  with  a  hori- 
zontal line  from  his  eye  to  the  pole? 

14.  Find  the  functions  of  45°,  60°,  30°,  0°,  90°,  before 
reading  the  next  article. 

5.  Trigonometric  functions  of  45°,  60°,  30°,  0°,  90°.  Varia- 
tion of  functions. 

J?  A.  Functions  of  45°.     Let 

AMP  be  an  isosceles  right- 
angled  triangle,  and  let  each 
of  the  sides  about  the  right 
angle  be  equal  to  a.  Then 
Fig.  1 1 .         angle  A  =  45°,  and  AP  =  a\/2 . 


sin  45°  =  sin  A  = 


MP 


1 


AP     aV2     V2* 
Thus,  by  definitions  Art.  3  and  Fig.  10, 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES 


sin  45°=-/-=,       tan  45°  =  1,  sec  45°=  V^ 

cos  45°  =  —=,       cot  45°  =  1 ,       cosec  45°  =  V2. 
V2 

The  sides  of  triangle  AMP  are  proportional  to  1,  1,  a/2. 

Hence,  in  order  to  produce  the  ratios  of  45°  quickly,  it  is 

merely  necessary  to  draw 

Fig.  11;    from  this  figure 

the  ratios  of    45°  can  be 

read  off  at  once. 

B.  Functions  of  30°  and 
60°.     Let  ABC  be  an  equi-     A°\ 
lateral  triangle.    From  any  -4r~    a      D 
vertex   B  draw  a  perpen- 
dicular BD  to  the  opposite  side  AC.    Then  angle  DAB  =  60°, 
angle  ABD  =  SQ°.  

If  AB=2a,  then  AD  =  a,  and  DB  =  V4a2-a2  =  ax/3. 

,.    Sin60°=sinZ)^=5|^^. 
AB       2a        2 

Thus,  from  Fig.  12, 

\/3 

sin  60° =-2",    tan60°  =  \/3,        sec  60° =2, 


cos  60° =4,      cot  60°=-  ~j=. 


2^ 

a/5' 


Also, 
Thus, 


sin30°=sm^D=^=|a4 


112 
sin  30°=-,       tan  30°=-^,        sec  30° =-7=, 
2'  V3  V3 

a/3  r 

cos  30°  -  — ,     cot  30°  -  V3,    cosec  30°  =  2. 


10 


ELEMENTS   OF  PLANE  TRIGONOMETRY 


In  ADB  the  sides  opposite  to  the  angles  30°,  60°,  90°,  are 
respectively  proportional  to  1,  V3,  2.  Hence,  in  order  to 
produce  the  functions  of  30°,  60°,  at  a  moment's  notice,  it 
is  merely  necessary  to  draw  Fig.  13,  from  which  these  func- 
tions can  be  immediately  read  off. 

C.  Functions  of  0°  and  90°.      Let 

the  hypotenuse  in  each  of  the  right- 
angled  triangles  in  Fig.  14  be  equal 
to  a. 

MP 

AP' 

AM 


sin  MAP 


cos  MAP" 


AP 


It  is  apparent  from  this  figure  that  if  the  angle  MAP 
approaches  zero,  then  the  perpendicular  MP  approaches 
zero,  and  the  hypotenuse  AP  approaches  to  an  equality  with 
AM;  so  that,  finally,  if  MAP  =  0,  then  MP  =  0,  and  AP  =  AM. 
Therefore,  when  MAP  =  0,  it  follows  that 


sin  0°=-=0,      tan  0°=-=0, 


cos0°=-  =  l,      sec0°=-  =  l. 

a  a 


(AM  approachmg  a\         :. 

Also  cot  0  =  I  ,Trt       t~t- --?{ )  =  unlimited  number  = 

\MP  approaching  0/ 


X 


approaching 

_Q     /     AP  equal  too    \         ,.    .    •, 

cosec  0  —  l  ttd r~' ?, )  =  unlimited  number  =  qo  . 

\MP  approaching  0/ 

As  MAP  approaches  90°,  AM  approaches  zero,  and  MP 


TRIGONOMETRIC  FUNCTIONS   OF  ACUTE  ANGLES         11 

approaches    to    an    equality    with    AP.    Therefore,    when 
MAP  =  90°,  it  follows  that 

sin90°  =  -  =  l,    tan90°  =  oo,  sec90°  =  oo, 

cos  90°=- =0,     cot  90°=-  =  0,    cosec90°=-  =  l. 

Thus  as  the  angle  increases  from  0°  to  90°,  its  sine 
increases  from  0  to  1;  its  cosine  decreases  from  1  to  0;  its 
tangent  increases  from  0  to  oo ;  its  cotangent  decreases  from 
oo  to  0;  its  secant  increases  from  1  to  oo  ;  its  cosecant  de- 
creases from  oo  to  1. 

EXAMPLES 

N.B.  Exponents.  In  trigonometry  (sin  x)n  is  usually  written 
sinn  x,  and  similarly  for  the  other  functions.  This  is  not  done 
when  n=  —1.  The  reason  for  this  exception  is  given  in  Art.  29. 
Find  the  numerical  value  of 

1.  sin  60° +  2  cos  45°.  2.  sec2  30°  + tan3  45°. 

3.  sin3  60°  +  cot3  30°.  4.  cos  0°  sin  45°  +  sin  90°  sec2  30°. 

5.  4  cos2  30°  sin2  60°  cos2  0°. 

6.  3  tan3  30°  sec3  60°  sin2  90°  tan2  45°.    7.  10  cos4  45°  sec6  30°. 

8.  2  sin5  30°  tan3  60°  cos3  0°. 

9.  x  cot3  45°  sec2  60°=  11  sin2  90°;  find  x. 

10.  x(cos30o  +  2sin90o+3cos45o-sin260o)  =  2sec0°-5sm90o; 
find  x. 


12 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


6.  Relations  between  the  trigonometric  functions  of  an 
acute  angle  and  those  of  its  comple- 
ment. When  two  angles  added  to- 
gether make  a  right  angle,  the  two 
angles  are  said  to  be  complementary, 
and  each  angle  is  called  the  comple- 
ment of  the  other. 

Thus,  in  Kg.  15,  P  =  90°-A  and  P 
is  the  complement  of  A.    Now 

MP 


Fig.  15. 


sin  ^==Tp"==  cos  P=  cos  (90°- A) 

AM 

cos  A  =-jTp=  sin  P  =  sin  (90°-  A) 

MP 

tan  A  =  -^  =  cot  P  =  cot  (90°-  A) 

cot  A  =  -^p  -  tan  P-  tan  (90°-  A) 

AP 
sec  A  =-7-Tr  =  cosec  P=  cosec  (90°—  A) 


cosec  A  — 


AP 
MP 


secP  =     sec  (90°-  A) 


These  six  relations  can  be  expressed  briefly: 
Each  trigonometric  function  of  an  angle  is  equal  to  the 
corresponding  co-function  of  its  complement. 


EXAMPLES 

1.  Compare  the  functions  of  30°  and  60°;   of  0°  and  90°. 

2.  Express  the  following  as  functions  of  angles  less  than  45°: 
sin  78°  20',  cos  80°  30',  tan  50°,  cot  65°,  sec  71°,  cosec  80°. 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES         13 

3.  If  ski  x=cos  (2z  +  40°),  find  a  value  of  x. 

4.  If  cot  2x = tan  (x—  30°),  find  a  value  of  x. 

6.  Show  that  in  a  triangle  ABC,  sin  JB  =  cos  i(A  +  C). 

7.  Relations  between  the  trigonometric  functions  of  an 
acute  angle. 

A.  Reciprocal  relations  between  the  functions. 
Inspection  of  the  definitions  (2),  Art.  3,  shows  that: 

(a)  sinA  = r.  cosec  A=— — r,  or,  sin  A  cosec  A=l: 

v  '  cosec  A  smi       '  ' 


(i) 


(b)  cos  A  = t,  sec  A= 7,  or,  cos  A  sec  -4=1; 

v  '  sec  A  cos  A7  ' 

1  1 

(c)  tan  A= — —r,         cot  A=- j,  or,  tan  A  cot  -4=1. 

cot  ^1  tan  ./L 

2?.  7%6  tangent  and  cotangent  in  terms  of  the  sine  and 
cosine. 

In  the  triangle  AMP  (Fig.  3), 

MP 
MP    AP     sin  A 
tanA~AM~AM  =  ^A;  (2) 

AP 
AM 


AM    AP     cos  A 
cot  A  =  MP=MP=stiA-  (3) 

AP 

C  Relations  between  the  squares  of  certain  functions. 
In  the  triangle  AMP  (Fig.  3),  indicating  by  WP2  the 
square  of  the  length  of  MP, 


MP  +AM  =AP  . 
On  dividing  each  member  of  this  equation  by  AP  ,  AM  , 
MP  ,  in  turn,  there  is  obtained 


14  ELEMENTS   OF  PLANE  TRIGONOMETRY 


(mp\2   (AMy   /Apy 

\APj  +\APj  ~\AP)  ' 


(mp\2  (AMy  /ap\ 
\am)  +\am)  ~\am) 

(mpv   /amv   /apv 

\MP, 


V     /AMy    (AP\ 
7  +\MPj  ~\MP) 


In  reference  to  the  angle  A,  these  equations  can  be 
written : 

sin2  A  +  cos2  A  =  l9 

tan2 .4  +  1         =sec2A,  (4) 

1  +  cot2  A  =  cosec2  A. 

Note.  An  equation  involving  trigonometric  functions  is  a 
trigonometric  equation.  Thus,  for  example,  tsaiA  =  l.  One 
angle  which  satisfies  this  equation  is  the  acute  angle  A  =  45°. 
Other  solutions  can  be  found  after  Art.  28  has  been  taken  up. 


EXAMPLES 

In  the  following  examples,  the  positive  values  of  the 
radicals  are  to  be  taken.  The  meaning  of  the  negative  values 
is  shown  in  Art.  20. 

1.  Given  that  sin  A  =  J,  find  the  other  trigonometric  ratios 
of  A  by  means  of  the  relations  shown  in  this  article. 

V3 


cosec  A  =— — r=2:     cos  A  =  \xl  —  sin2  A=— ~ : 
sin  A       '  2    3 


sec  A  = 


1           2 

sin  A       1 
tan  A—          .—     ,_ ; 
cos  A     V3 

cos  A     V3 ' 

cot  A  =  — 

\=V3. 

tan  A 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES         15 

These   results   may   be   verified   by   the   method   used   in 
solving  Exs.  1,  5-7/ Art.  4. 

2.  Express  all  the  ratios  of  angle  A  in  terms  of  sin  A. 


sin  A  =  sin  A;  cos  A  =  v. 1  —  sin2  A ; 


sin  A  sin  A  1         V 1  —sin2  A 

tanA  = r=     , :  cotA  =  - -  = : — ; 

cos  A     Vl  —sin2  A  tan  A  sin  A 

sec^l  = -:-=     . —    .         ;  cosecA  =  — 


cosA     Vl  —sin2  A '  sin  A 


3.  Prove  that    ; ; — t  +  t-— : — 7=  2  sec2  A. 

1—  sin  A     1  +  smA 

1  _1 _2 2 

1  — sin  A     1  +  sinA     1  —  sin2  A      cos2  A 


4.  Prove  that      sec4  A  -1  =  2  tan2  A+tan4  A. 
sec4A-l=(sec2A)2-l  =  (l+tan2A)2-l  =  2tan2A+tan4A. 

5.  Solve  the  equation  4  sin  6—3  cosec  0=0. 

3 
4sin0 — =— z=0. 
sin  # 

.*.     4  sin2  0-3  =  0. 

3  V3  .  V3 

.*.     sin2  0= j,     .*.     sin^=+-^-,     and     sin  0= —. 

On  taking  the  pZws  sign,  one  solution  is  the  acute  angle 
0=60°;  other  solutions  will  be  found  later.  For  the  minus 
sign  there  is  also  a  set  of  solutions;  these  will  be  found  later. 


16  ELEMENTS  OF  PLANE  TRIGONOMETRY 

6.  Solve      2  sin2  0  cosec  0-5  +  2  cosec  0=0, 

2  sin2  0  2 

sin  0  sin  0 

2  sin2  0-5  sin  0  +  2  =  0, 

(2sin0-l)(sin0-2)=O. 

.*.     sin0=J,     and     sin  0=2. 

The  acute  angle  whose  sine  is  \  is  30°;  hence  0=30°  is 
one  solution.  The  sine  cannot  exceed  unity;  hence  sin  0=2 
does  not  afford  any  solution. 

7.  Given  cos  A  =  j,  sin  B= J,  tan  C=2,  cot  D  =  %,  sec  E=S, 
cosec  ^=2.5;  find  the  other  trigonometric  ratios  of  A,  B,  C, 
D,  E,  F,  by  the  algebraic  method.  Verify  the  results  by  the 
method  used  in  Art.  4. 

8.  Find  by  the  algebraic  method  the  ratios  required  in 
Exs.  1,  5-7,  10,  11,  Art.  4. 

9.  Express  all  the  trigonometric  ratios  of  an  angle  A  in 
terms  of:  (a)  cos  A;  (6)  tan  A)  (c)  cot  A;  (d)  sec  A;  (e) 
cosec  A.  Arrange  the  results  and  those  of  Ex.  2  neatly  in 
tabular  form. 

Prove  the  following  identities: 

10.  (sec2  A  —  1)  cot2  A  =  1 ;  cos  A  tan  A  =  sin  A ; 
(1  —  sin2  A)  sec2  A  =  l. 

11.  sin2  0  sec2  0 = sec2  0-1;  tan2  0- cot2  0= sec2  0- cosec2  0, 


sin  A       cos  A 
)sec  A     ; 

l  +  sin0 


sec2  A     cosec2  A       '     cosec  A     sec  A. 


(tan  0+ sec  0)2  = 


1  -sin  0' 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES  17 

13.  sec2  A  +  cosec2  A  =  tan2  A  +  cot2  A  +  2 ; 

1  +  tan2  A     sin2  A  cosec  A 


1  +  cot2  A     cos2  A '     cot  A  +  tan  A 


cos  A. 


cos  A  sin  A 

14.  - — -r+: — r=sin  A+cos  A; 

1  —tan  A     1  —cot  A 


1 


■=  sec  A  +  tan  A ;    sec4  A  —sec2  A  =tan4A  +tan2A. 


sec  A  —tan  A 

Solve  the  following  equations: 

15.  2  sin  0=2 -cos  6.  16.  tan  0  +  cot  0=2. 

17.  tan  0  +  3  cot  0  =  4.  18.  6  sec2  0-13  sec  0+5  =  0. 

19.  8  sin2  0-10  sin  0+3  =  0.      20.  sin  0  +  2  cos  0=2.2. 


CHAPTER  II 
SOLUTION   OF  RIGHT-ANGLED  TRIANGLES 

Applications 

8.  Solution  of  a  triangle.  There  are  two  methods  for 
finding  the  unknown  parts  of  triangle  (sides  and  angles) 
when  a  sufficient  number  of  parts  are  given,  viz. : 

(a)  The  graphical  method; 

(6)  The  method  of  computation. 

The  graphical  method  consists  in  drawing  a  triangle  which 
has  angles  equal  to  the  given  angles,  and  sides  proportional 
to  the  given  sides,  and  then  measuring  the  remaining  parts 
directly  from  the  drawing. 

The  method  of  computation  consists  in  writing  the  formulas 
by  which  the  unknown  parts  can  be  found  and  doing  the 
arithmetical  work. 

The  latter  method  is  the  more  exact.  The  graphical 
method  affords  a  rough  check  on  the  results  obtained  by 
computation  and  leads  to  the  detection  of  large  errors. 

General  suggestions  for  solving  problems. 

(1)  Make  an  off-hand  estimate  as  to  what  the  magnitude 
required  may  be,  and  write  this  estimate  down; 

(2)  Solve  the  problem  by  the  graphical  method; 

(3)  Solve  the  problem  by  the  method  of  computation; 

(4)  Check  the   accuracy   of   the   results   arithmetically, 

using  formulas  not  used  in  process  (3). 

18 


SOLUTION  OF  RIGHT-ANGLED  TRIANGLES  19 

9.  Cases  in  the  solution  of  right-angled  triangles.     All  the 

possible  sets  of  two  elements  that  can  be  made  from  the 
three  sides  and  the  two  acute  angles  of  a  right-angled  triangle 
are  the  following: 

(1)  The  two  sides  about  the  right  angle. 

(2)  The  hypotenuse  and  one  of  the  sides  about  the  right 
angle. 

(3)  The  hypotenuse  and  an  acute  angle. 

(4)  One  of  the  sides  about  the  right  angle,  and  an  acute 
angle. 

(5)  The  two  acute  angles.  (An  unlimited  number  of 
triangles  can  have  two  given  acute  angles.) 

Relations  (2),  Art.  3,  for  the  triangle  AMP  (Fig.  3)  show 
that  if  any  two  sides  of  a  right-angled  triangle  be  given,  or 
any  side  and  an  acute  angle  be  given,  the  remaining  parts  of 
the  triangle  can  be  found. 

In  solving  a  triangle,  the  general  method  of  procedure, 
after  making  an  off-hand  estimate  and  finding  an  approxi- 
mate solution  by  the  graphical  method,  is  as  follows: 

First :  Write  all  the  relations  (or  formulas)  which  are  to  be 
used  in  solving  the  problem. 

Second :  Write  the  check  formulas. 

Third:  In  making  the  computations  arrange  the  work  as 
neatly  as  possible. 

This  last  is  important,  because,  by  attention  to  this  rule, 
the  work  is  presented  clearly,  and  mistakes  are  less  likely  to 
occur.  The  computations  may  be  made  either  with  or  with- 
out the  help  of  logarithms.  The  calculations  can  generally 
be  made  more  easily  and  quickly  by  using  logarithms. 


20  ELEMENTS   OF  PLANE  TRIGONOMETRY 


EXAMPLES 

1.  In   the    triangle    ABC,    right-angled    at    C,    a=42  ft., 
6=56  ft.     Find  the  hypotenuse   and  the 
acute  angles. 

I.    Computation     without      logarithms. 
[Four-place  tables.] 

a     42 
tanA  =  r  =  —  =  .7500.    .*.  A  =  36°  52'.2. 
o     56 

B=90-A.  -      .'.  £=53°7'.8. 


c=Va2  +  62  =  V1764  +  3136.  .'.    c=70  ft. 

Check:       a=c  cos  B = 70 X cos  53°  7'.8=70X.6000  =  42  ft. 

II.  Computation  with  logarithms. 

Given:  a=42  ft.  To  find:  *    A  = 

6=56  ft,  B= 

a  c= 

Formulas:  tan  A  =  ^.  (1) 

5=90° -A.     (2)    ^ecfcs;  tan  £=-. 
,  a  a2  =  c2-b2 

C_sinA'  «(c+6)(c-fc). 

Logarithmic  formulas:  log  tan  A  =  log  a  —log  6. 

log  c = log  a  —log  sin  A . 

log  a  =  1 .62325  log  a  =  1 .62325 

log  6=1.74819  log  sin  .4  =  9.77815-10 

.*.    log  tan  A  =  9.87506  -10  .\    log  c=  1.84510 

.*.     A  =  36°  52'  12"  .'.     c=70 

.'.     5=53°  T  48" 

*  This  is  to  be  filled  after  the  values  of  the  unknown  quantities  have 
been  found.  It  is  advisable  to  indicate  the  given  parts  and  the  unknown 
parts  clearly. 


SOLUTION  OF  RIGHT-ANGLED  TRIANGLES  21 

The  work  can  be  more  compactly  arranged,  as  follows: 

Checks: 
log  a=  1.62325  log  tan  B=  10.12494-10 

log  6=1.74819  /.     B    =  53°  7'  48" 

,\     log  tan  A  =  9.87506 -10  c  +  6=126 

.'.    A  =  36°52'  12"  c-6=    14 

/.     5  =  53°  T  48"  log  (c  +  6)  =  2.10037 

log  sin  A  =  9.77815 -10  log  (c  -6)  =  1.14613 

.*.    log  c=  1.84510  .      ,        2     0  0/fncn 

&    _7n  ..     loga2  =  3.24650 

*''     °  ;.     log  a  =1.62325 

Note.  In  every  example  it  is  advisable  to  make  a  complete 
skeleton  scheme  of  the  solution,  before  using  the  tables  and 
proceeding  with  the  actual  computation.  In  this  exercise,  for 
instance,  such  a  skeleton  scheme  can  be  seen  on  erasing  all 
the  numerical  quantities  in  the  equations  that  follow  the 
logarithmic  formulas. 

2.  In  a  triangle  ABC  right  angled  at  C,  c=60  ft.,  6=50  ft.; 
find  side  a  and  the  acute  angles. 

I.  Computation  without  logarithms. 


6     50  *   o 

cos  A  =  -  =—=.8333.     .'.     A=33°33'.75. 
c     60 

B=90°-A.  .'.     5=56°26'.25.     A        &-»**•  G 

a=c  sin  A  =  60X.5528       =33.17  ft.  FlG"  17' 

Check:   a=6  tan  A  =  50 X. 6635= 33.17. 

II.  Computation  with  logarithms. 

Given:  c=60  ft.  .  To  find:  A  = 

6=50  ft.  B= 

6  <*= 

c'  Checks:  a2  =  c2— 62=(c+6)(c— 6). 
B=90°-A.  a=6tanA. 

a  =  c  sin  A. 


22 


ELEMENTS   OF  PLANE  TRIGONOMETRY 


Logarithmic  formulas :    log  cos  A  =  log  b  —log  c. 
(If  necessary.)  log  a  =  log  c  +  log  sin  A . 


log  6=  1.69897 
log  c=  1.77815 


(1) 
(2) 


log  cos  A  =  9.92082 -10  (3) 
=  (D-(2) 
/.     A  =  33°  33'  27" 
.-.     £=56p26'  33" 
log  sin  A  =  9.74255 -10  (4) 
.'.     log  a=  1.52070  (5) 

=  (2) +  (4) 
.*.     a  =  33.16 


log  tan  A  =  9.82173 - 
.*.  log  a=  1.52070 

=  (l)  +  (6) 

c+6=110 
c-b  =    10 

log  (c+6)  =  2.04139 

log(c-6)  =  l 

.-.     log  a2= 3.04139 

.'.     log  a  =1.52070 


•10  (6) 
(7) 


Note.  There  is  a  slight  difference  between  the  results 
obtained  by  the  two  methods.  This  is  due  to  the  fact  that 
the  calculations  have  been  made  with  a  four-place  table  in 
one  case,  and  with  a  five-place  table  in  the  other.  A  four- 
place  table  will  give  an  angle  correctly  to  within  one  minute; 
a  five-place  table  will  give  it  correctly  to  within  six  seconds, 
and  sometimes  to  within  a  second. 

3.  In  a  triangle  right  angled  at  C,  the  hypotenuse  is  250  ft., 
and  angle  A  is  67°  30r.     Solve  the  triangle. 

I.  Computation  without  logarithms. 
B  =  90°  -A  =  90°  -67°  30r  =  22°  30'. 

a  =  c  sin  A  =  250  Xsin  67°  30,  =  250X. 9239=230.98. 
6  =  ccosA  =  250Xcos67°30,  =  250X.3827=  95.68. 
Checks:  a2 =c2—b2,     or     a=6tanA. 

II.  Computation  with  logarithms. 
Given  c=250  ft.  To  find:   B= 

A  =  67°  30'.  a= 

b= 
Formulas:   B  =  90° -A.  Checks:    a2=c2-b2 

a  =  c  sin  A.  =(c+6)(c— 6). 

6  =  ccosA. 


SOLUTION  OF   RIGHT-ANGLED  TRIANGLES 


23 


Logarithmic  formulas: 

.*.     B=22°  30' 
log  c= 2.39794 
log  sin  A  =  9.96562  -10 
log  cos  A  =  9.58284  -10 
.-.    log  a  =  2.36356 
.-.     log  6=  1.98078 
.'.     a=230.97 
.'.     6=   95.67 


log  a=log  c+log  sin  A. 
log  6= log  c+log  cos  A. 

c  + 6  =  345.67 
c -6=  154.33 
log  (c  +  6)  =  2.53866 
log  (c -6)  =  2.18845 
.*.     loga2=4.72711 
.'.     log  a  =2.36356 


4.  In  a  triangle  ABC  right  angled  at  C,  6=300  ft.  and 
A  =  37°  20'.     Soive  the  triangle. 

I.  Computation  without  logarithms. 
B=  90°  -A  =  90°  -37°  30'  =  52°  40'. 
6  300 


c= 


=377.3. 


cos  A     .7951 
a=b  tan  A  =  300  X. 7627  =  228.8. 
Checks :  a2  =  c2—b2,     a  =  c  sin  A . 
II.  Computation  with  logarithms. 


Given 


Formulas: 


A  =  37°  20'. 
6=300  ft. 
B= 90°  -A. 
6 


cos  A 
a=6  tan  A. 

.*.     £=52°  40' 
log  6=2.47712 
log  cos  A  =  9.90043 -10 
log  tan  A  =  9.88236 -10 
.'.     logc=2.57669 
.'.     loga  =  2.35948 
.'.     c=377.3 
.-.     a=228.8 


To  find:    B= 

c= 

a~ 
Checks :    a2=c2—b2 

=  (c+6)(c-6) 

c+6  =  677.3 
c-6=  77.3 
log  (c+6)  =  2.83078 
log  (c -6)  =  1.88818 
.*.     loga2  =  4.71896 
.'.     log  a  =2.35948 


24  ELEMENTS  OF  PLANE  TRIGONOMETRY 

N.B.  Check  all  results  in  the  following  examples.  The 
given  elements  belong  to  a  triangle  ABC  which  is  right  angled 
at  C. 

From  the  given  elements  solve  the  following  triangles: 

5.  c=18.7,     a=  16.98.  6.  a=  194.5,    6  =  233.5. 

7.  c=2934,  ,4  =  31°  14'  12".  8.  a=36.5,     J3=68°52'. 

9.  a  =  58.5,     6=100.5.  10.  c=45.96,    a=  1.095. 

11.  c=324,     A  =  48°17'.  12.  6  =  250,      ^  =  51°  19'. 

13.  c=1716,  ^  =  37°  20' 30".  14.  a  =  2314,     6=1768. 

15.  6  =  3741,  A  =  27°45'20".  16.  c  =  50.13,    a  =  24.62. 

Solve  Exs.  17-24  by  two  methods,  viz. :  (1)  with  logarithms; 

(2)  without  logarithms. 


17. 

a =40, 

5=62°  40'. 

18. 

c=9, 

a  =5. 

19. 

a =4.5, 

6=7.5. 

20. 

c=15, 

,4  =  39°  40' 

21. 

c=12, 

£=71°  20'. 

22. 

c=12, 

a=8. 

23. 

6=15, 

5=42°  30'. 

24. 

a=8, 

6=12. 

10.  Projection  of  a  straight  line  upon  another  straight  line. 
Let  AB,  of  length  I,  be  inclined 
at  an  angle  a  to  LR.     If  per- 
pendiculars AM,  BN,  are  drawn 
to  LR,  MN  is  called  the  (orthog- 


i/> 


onal)  projection   of   AB  on  LR.  is N 

Through  A  draw  AD  parallel  FlG  20 

to  LR.    Then 

Projection  =  MN  =  AD  =  AB  cos  DAB  =  I  cos   a. 

That  is,  the  projection  of  one  straight  line  upon  another 
straight  line  is  equal  to  the  product  of  the  length  of  the  first 
line  and  the  cosine  of  the  angle  of  inclination  of  the  two  lines. 


SOLUTION   OF  RIGHT-ANGLED  TRIANGLVES  25 


EXAMPLES 

In  working  these  examples  use  logarithms  or  not,  as  appears 
most  convenient.     Check  the  results. 

1.  A  ladder  28  ft.  long  is  leaning  against  the  side  of  a  house, 
and  makes  an  angle  27°  with  the  wall.  Find  its  projections 
upon  the  wall  and  upon  the  ground. 

2.  What  is  the  projection  of  a  line  87  in.  long  upon  a  line 
inclined  to  it  at  an  angle  47°  30'? 

3.  What  are  the  projections:  (a)  of  a  line  10  in.  long  upon 
a  line  inclined  22°  30'  to  it?  (i)  of  a  line  27  ft.  6  in.  long  upon 
a  line  inclined  37°  to  it?  (c)  of  a  line  43  ft.  7  in.  long  upon  a 
line  inclined  67°  20'  to  it?  (d)  of  a  line  34  ft.  4  in.  long  upon  a 
line  inclined  55°  47'  to  it? 

11.  Measurement  of  heights  and  distances.  When  an 
object  is  above  the  observer's  eye,  the  angle  between  the  line 
from  the  eye  to  the  object,  and  the  horizontal  line  through 


Fig.  21. 

the  eye  and  in  the  same  vertical  plane  as  the  first  line,  is 
called  the  angle  of  elevation  of  the  object,  or  simply  the 
elevation  of  the  object.  When  the  object  is  below  the 
observer's  eye,  this  angle  is  called  the  angle  of  depression 
of  the  object,  or  simply  the  depression  of  the  object. 


26 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


EXAMPLES 

1.  At  a  point  150  ft.  from,  and  on  a 
level  with,  the  base  of  a  tower,  the  angle 
of  elevation  of  the  top  of  the  tower  is 
observed  to  be  60°.  Find  the  height  of 
the  tower. 

Let  AB  be  the  tower,  and  P  the  point  of 
observation. 

By  the  observations, 

AP=150ft.,       APB=W°. 


Fig.  22. 


AB=AP  tan  60°=  150X^3  =  150X1.7321  =  259.82  ft. 

2.  In  order  to  find  the  height  of  a  hill,  a  line  was  measured 
equal  to  100  ft.,  in  the  same  level  with  the  base  of  the  hill, 
and  in  the  same  vertical  plane  with  its  top.  At  the  ends  of 
this  line  the  angles  of  elevation  of  the  top  of  the  hill  were 
30°  and  45°.     Find  the  height  of  the  hill. 

Let  P  be  the  top  of  the  hill,  and  AB  the  base  line.  The 
vertical  line  through P  will  meet,  AB  produced  in C.  AB=  100  ft. 

CAP=S0°,  CPP=45°;  the  height 
CP  is  required.  Let  BC=x,  and 
CP=y.  [ 

In  triangle  CAP, 

CP 


Fig.  23. 


in  CBP% 


2^=tan30°; 
CP 


BC 


=  tan  45°. 


Hence, 


V 


z+100 


=  tan  30°  =.57735, 


and 


-=tan45°=l. 

x 


a) 

(2) 


From  (2),  x=y.     Substitution  in  (1)  gives 
--(»/+•  100)  X. 57735. 


SOLUTION  OF  RIGHT-ANGLED   TRIANGLES  27 

.\    2/(1 -.57735)  =57.735. 

3.  A  flagstaff  30  ft.  high  stands  on  the  top  of  a  cliff,  and 
from  a  point  on  a  level  with  the  base  of  the  cliff  the  angles  of 
elevation  of  the  top  and  bottom  of  the  flagstaff  are  observed 
to  be  40°  20'  and  38°  20',  respectively.  Find  the  height  of 
the  cliff. 

Let  BP  be  the  flagstaff  on  the  top  of  the  cliff  BL,  and  let 
C  be  the  place  of  observation.  PP=30  ft.,  LCB=S8°  20', 
LCP=40°  20'.     Let  CL  =  x,  LB=y. 


P        7 

/    J30J.FY. 

In  LCB, 

LB 

y^=tan38°20r; 

/X0^ 

i.e., 

-=.7907. 

In  LCP, 

LP 

-^=tan40°20'; 

J4. QQ +,Jj 

Fig.  24. 

i.e., 

y+30-.8491. 

Hence,  on  division, 

y        7907 

y 

+  30     8491' 

On  solving  for  y, 

LB=y=406.18ft. 

4.  At  a  point  180  ft.  from  a  tower,  and  on  a  level  with  its 
base,  the  elevation  of  the  top  of  the  tower  is  found  to  be 
65°  40/5.     What  is  the  height  of  the  tower? 

5.  From  the  top  of  a  tower  120  ft.  high  the  angle  of 
depression  of  an  object  on  a  level  with  the  base  of  the  tower 
is  27°  43'.  What  is  the  distance  of  the  object  from  the  top 
and  bottom  of  the  tower? 


28  ELEMENTS  OF  PLANE  TRIGONOMETRY 

6.  From  the  foot  of  a  post  the  elevation  of  the  top  of  a 
column  is  45°,  and  from  the  top  of  the  post,  which  is  27  ft. 
high,  the  elevation  is  30°.  Find  the  height  and  distance  of 
the  column. 

7.  From  the  top  of  a  cliff  120  ft.  high  the  angles  of  depres- 
sion of  two  boats,  which  are  due  south  of  the  observer,  are 
20°  20'  and  68°  40'.     Find  the  distance  between  the  boats. 

8.  From  the  top  of  a  hill  450  ft.  high,  the  angle  of  depres- 
sion Of  the  top  of  a  tower,  which  is  known  to  be  200  ft.  high, 
is  63°  20'.  What  is  the  distance  from  the  foot  of  the  tower 
to  the  top  of  the  hill? 

9.  From  the  top  of  a  hill  the  angles  of  depression  of  two 
consecutive  mile-stones,  which  are  in  a  direction  due  east,  are 
21°  30'  and  47°  40'.     How  high  is  the  hill? 

10.  For  an  observer  standing  on  the  bank  of  a  river,  the 
angular  elevation  of  the  top  of  a  tree  on  the  opposite  bank  is 
60°;  when  he  retires  100  ft.  from  the  edge  of  the  river  the 
angle  of  elevation  is  30°.  Find  the  height  of  the  tree  and  the 
breadth  of  the  river. 

11.  Find  the  distance  in  space  travelled  in  an  hour,  in  con- 
sequence of  the  earth's  rotation,  by  an  object  in  latitude 
44°  20'.     [Take  earth's  diameter  equal  to  8000  mi.] 

12.  At  a  point  straight  in  front  of  one  corner  of  a  house, 
its  height  subtends  an  angle  34°  45',  and  its  length  subtends 
an  angle  72°  30';  the  height  of  the  house  is  48  ft.  Find  its 
length. 

12.  Solution  of  isosceles  triangles.  An  isosceles  triangle 
can  often  be  solved  on  dividing  it  into  two  equal  right- 
angled  triangles. 


SOLUTION  OF  RIGHT-ANGLED  TRIANGLES  29 

EXAMPLES 

1.  The  base  of  an  isosceles  triangle  is  24  in.  long,  and  the 
vertical  angle  is  48°;    find  the  other  angles  n 
and  sides,  the  perpendicular  from  the  vertex 
and  the  area.     Only  the  steps  in  the  solution 
will  be  indicated.     Let  ABC  be  an  isosceles 
triangle  having  base  Ai?=24  in.,  angle  (7=48°.           / 
Draw  CD  at  right  angles  to  base;    then  CD        I 
bisects  the  angle  ACB  and  base  A B.     Hence,    .Ajf-^. 
in    the    right-angled    triangle    ADC,    AD=           Fig  25 
iAB=12,  ACD=iACB=24°.     Hence,  angle 
A,    sides    AC,    DC,    and   the   area,    can   be   found. 

2.  In  an  isosceles  triangle  each  of  the  equal  sides  is  363  ft., 
and  each  of  the  equal  angles  is  75°.  Find  the  base,  per- 
pendicular on  base,  and  the  area. 

3.  In  an  isosceles  triangle  each  of  the  equal  sides  is  241  ft., 
and  their  included  angle  is  96°.  Find  the  base,  angles  at  the 
base,  height,  and  area. 

4.  In  an  isosceles  triangle  the  base  is  65  ft.,  and  each  of  the 
other  sides  is  90  ft.     Find  the  angles,  height,  and  area. 

5.  In  an  isosceles  triangle  the  base  is  40  ft.,  height  is  30  it. 
Find  sides,  angles,  area. 

6.  In  an  isosceles  triangle  the  height  is  60  ft.,  one  of  equal 
sides  is  80  ft.     Find  base,  angles,  area. 

7.  In  an  isosceles  triangle  the  height  is  40  ft.,  each  of  equal 
angles  is  63°.     Find  sides  and  area. 

8.  In  an' isosceles  triangle  the  height  is  63  ft.,  vertical  angle 
is  75°.     Find  sides  and  area. 

13.  Related    regular    polygons    and    circles.     Among  the 


30 


ELEMENTS   OF  PLANE  TRIGONOMETRY 


problems  on  regular  polygons  which  can  be  solved  with  the 
help  of  right-angled  triangles  are  the  following: 

(a)  Given  the  length  of  the  side  of  a  regular  polygon  of 
a  given  number  of  sides,  to  find  its  area;  also,  to  find  the 
radii  of  the  inscribed  and  circumscribing  circles  of  the 
polygon; 

(6)  To  find  the  length  of  the  sides  of  regular  polygons  of 
a  given  number  of  sides  which  are  inscribed  in,  and  circum- 
scribed about,  a  circle  of  given  radius. 

For  example,  in  Fig.  26,  let  AB  be  a  side,  equal  to  2a,  of 

a  regular  polygon  of  n  sides,  and  let  C  be  the  centre  of  the 

inscribed  circle.    Draw  CA,  CB,  and  draw  CD  at  right  angles 

to  AB-    Then  D  is  the  middle  point  of  AB. 

360°    180° 
By  geometry,  angle  ACD  =  %ACB  =  h 

Also,  angle  DAC= 90°-  ACT). 


n 


n 


Thus,  in  the  triangle  ADC,  the  side  AD  and  the  angles  are 
known;  therefore  CD,  the  radius  of  the  circle  inscribed  in  the 
polygon,  can  be  found.  On  making  similar  constructions,  the 
solution  of  the  other  problems  referred  to  above  will  be 


Fig. 


apparent.  The  perpendicular  from  the  centre  of  the  circle 
to  a  side  of  the  inscribed  polygon  is  called  the  apothem  of 
the  polygon. 


SOLUTION  OF  RIGHT-ANGLED  TRIANGLES  31 


EXAMPLES 

1.  The  side  of  a  regular  heptagon  is  14  ft. :  find  the  radii 
of  the  inscribed  and  circumscribing  circles;  also,  find  the 
difference  between  the  areas  of  the  heptagon  and  the  inscribed 
circle,  and  the  difference  between  the  area  of  the  heptagon 
and  the  area  of  the  circumscribing  circle. 

2.  The  radius  of  a  circle  is  24  ft.  Find  the  lengths  of  the 
sides  and  apothems  of  the  inscribed  regular  triangle,  quadri- 
lateral, pentagon,  hexagon,  heptagon,  and  octagon.  Compare 
the  area  of  the  circle  and  the  areas  of  these  regular  polygons; 
also  compare  the  perimeters  of  the  polygons  and  the  circum- 
ference of  the  circle. 

3.  For  the  same  circle  as  in  Ex.  2,  find  the  lengths  of  the 
sides  of  the  circumscribing  regular  figures  named  in  Ex.  2. 
Compare  their  areas  and  perimeters  with  the  area  and  cir- 
cumference of  the  circle. 

4.  If  a  be  the  side  of  a  regular  polygon  of  n  sides,  show 
that  R,  the  radius  of  the  circumscribing  circle,  is  equal  to 

180° 

iacosee ;    and  that  r,  the  radius  of  the  circle  inscribed, 

n 

180° 
is  equal  to  \a  cot -. 

5.  If  r  be  the  radius  of  a  circle,  show  that  the  side  of  the 

•    180°  ,     , 

regular  inscribed  polygon  of  n  sides  is  2r  sin ;    and  that 

180° 


the  side  of  the  regular  circumscribing  polygon  is  2r  tan 

f» 

6.  If  a  be  the  side  of  a  regular  polygon  of  n  sides.  R  the 

radius  of  the  circumscribing  circle,  and  r  the  radius  of  the 

180° 
circle    inscribed,    show   that    area    of    polygon =Jna2  cot 

.   .    360°        Q         180° 

=  hnR2  sin =  nrz  tan . 

2  n  n 


32 


.ELEMENTS  OF  PLANE  TRIGONOMETRY 


14.  Problems  requiring  a  knowledge  of  the  points  of  the 
Mariner's  Compass.  The  circle  in  the  Mariner's  Compass  is 
divided  into  32  equal  parts,  each  part  being  thus  equal  to 
360°  ■*■ 32,  i.e.,  11J°.  The  points  of  division  are  named  as 
indicated  on  the  figure. 


Fig.  27. 


It  will  be  observed  that  the  points  are  named  with  refer- 
ence to  the  points  North,  South,  East,  and  West,  which  are 
called  the  cardinal  points.  Direction  is  indicated  in  a  variety 
of  ways.  For  instance,  suppose  C  were  the  centre  of  the 
circle;  then  the  point  P  in  the  figure  is  said  to  bear  E.N.E. 
from  C,  or,  from  C  the  bearing  of  P  is  E.N.E.  Similarly, 
C  bears  W.S.W.  from  P,  or,  the  bearing  of  C  from  P  is 
W.S.W.  The  point  E.N.E.  is  2  points  North  of  East,  and 
6  points  East  of  North.  Accordingly,  the  phrases  E.  22J°  N., 
N.  67J°  E.,  are  sometimes  used  instead  of  E.N.E. 


s 

Fig.  28. 

If 

LA  is  6  mi., 

SOLUTION  OF  RIGHT-ANGLED  TRIANGLES  33 


EXAMPLES 

1.  Two  ships  leave  the  same  dock  at  8  a.m.  in  directions 
S.W.  by  S.,  and  S.E.   by  E.  at  rates 
of  9  and  9^  mi.  an  hour  respectively. 
Find   their    distance   apart,    and   the 
bearing  of  one  from  the  other  at  10  /& 
a.m.  and  at  noon. 

2.  From  a  lighthouse  L  two  ships    *?(£"-'-- 
A  and  B  are  observed  in  a  direction 
N.E.    and    N.    20°  W.    respectively. 

At  the  same  time  A  bears  S.E.  from  B. 
what  is  LB1 

15.  Examples  in  the  measurement  of  land.  In  order  to 
find  the  area  of  a  piece  of  ground,  a  surveyor  measures 
distances  and  angles  sufficient  to  provide  data  for  the  com- 
putation. An  account  of  his  method  of  doing  this  belongs 
to  works  on  surveying.  This  article  merely  gives  some 
examples  which  can  be  solved  without  any  knowledge  of 
professional  details.  In  solving  these  problems,  the  student 
should  make  the  plotting  or  mapping  an  important  feature 
of  his  work. 

The  Gunter's  chain  is  generally  used  in  measuring  land. 
It  is  4  rods  or  66  feet  in  length,  and  is  divided  into  100  links. 

An  acre  =  10  square  chains  =  4  roods  =  160  square  rods  or 
poles. 

EXAMPLES 

1.  A  surveyor  starting  from  a  point  A  runs  S.  70°  E.  20 
chains,  thence  N.  10°  W.  20  chains,  thence  N.  70°  W.  10  chains, 
thence  S.  20°  W.  17.32  chains  to  the  place  of  beginning. 
What  is  the  area  of  the  field  which  he  has  gone  around? 


34 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


Fig.  29. 


Make  a  plot  or  map  of  the  field,  namely,  ABCD.  Here 
A  B  represents  20  chains,  and  the  bearing  of  B  from  A  is  S. 
70°  E.  BC  represents  20  chains,  and 
the  bearing  C  from  B  is  N.  10°  W., 
and  so  on.  Through  the  most  west- 
erly point  of  the  field  draw  a  north- 
and-south  line.  This  line  is  called 
the  meridian.  In  the  case  of  each  line 
measured,  find  the  distance  that  one 
end  of  the  line  is  east  or  west  from  the 
other  end.  This  easting  or  westing 
is  called  the  departure  of  the  line.  Also 
find  the  distance  that  one  end  of  the 
line  is  north  or  south  of  the  other  end. 
This  northing  or  southing  is  called  the 
latitude  of  the  line.  For  example,  in 
Fig.  29,  the  departures  of  AB,  BC,  CD,  DA,  are  BXB,  BL} 
CH,  DD\,  respectively;  the  latitudes  of  the  boundary  lines 
are  ABX,  BiCi,  CiDh  DXA,  respectively.  The  following 
formulas  are  easily  deduced: 

Departure  of  a  line = length  of  line X  sine  of  the  bearing; 
Latitude  of  a  line = length  of  line  X  cosine  of  the  bearing. 

By  means  of  the  departures,  the  meridian  distance  of  a 
point  (i.e.,  its  distance  from  the  north-and-south  line)  can  be 
found.  Thus  the  meridian  distance  of  C  is  C\C,  and  C\C= 
DtD  +  HC.  Hence  in  Fig.  2.9,  ABU  BXB,  BXCU  dC,  CXDU 
D\D  can  be  computed.     Now 

area  ABCD  =  trapezoid  D1DCC1  +  trapezoid  C\CBBi 
—triangle  ADD\        —triangle  ABB%m 


The  areas  in  the  second  member  can  be  computed;  it  will 
be  found  that  area  ABCD  =  26  acres. 

Note.  Sometimes  the  bearing  and  length  of  one  of  the 
lines  enclosing  the  area  is  also  required.  These  can  be  com- 
puted by  means  of  the  latitudes  and  departures  of  the  given 
lines. 


SOLUTION  OF  RIGHT-ANGLED  TRIANGLES  35 

2.  In  Ex.  1,  deduce  the  length  and  bearing  of  DA  from  the 
lengths  and  bearings  of  AB,  BC,  CD. 

3.  A  surveyor  starts  from  A  and  runs  4  chains  S.  45°  E. 
to  B,  thence  5  chains  E.  to  C,  thence  6  chains  N.  40°  E.  to  D. 
Find  the  distance  and  bearing  of  A  from  D;  also,  the  area  of 
the  field  ABCD.  Verify  the  results  by  going  around  the  field 
in  the  reverse  direction,  and  calculating  the  length  and  bear- 
ing of  BA  from  the  lengths  and  directions  of  AD,  DC,  CB. 

4.  A  surveyor  starts  from  one  corner  of  a  pentagonal  field, 
and  runs  N.  25°  E.  433  ft.,  thence  N.  76°  55'  E.  191  ft.,  thence 
S.  6°  41'  W.  539  ft.,  thence  S.  25°  W.  40  ft.,  thence  N.  65°  W. 
320  ft.  Find  the  area  of  the  field.  Deduce  the  length  and 
direction  of  one  of  the  sides  from  the  lengths  and  directions  of 
the  other  four. 

5.  From  a  station  within  a  hexagonal  field  the  distances  of 
each  of  its  corners  were  measured,  and  also  their  bearings; 
required  its  plan  and  area,  the  distances  in  chains  and  the 
bearings  of  the  corners  being  as  follows:  7.08  N.E.,  9.57  N. 
J  E.,  7.83  N.W.  by  W.,  8.25  S.W.  by  S.,  4.06  S.S.E.  7°  E., 
5.89  E.  by  S.  3^°  E. 


CHAPTER  III 

ANGLES  IN  GENERAL  AND  THEIR  TRIGONOMETRIC 
FUNCTIONS 

16.  General  definition  of  an  angle.  The  amount  of  rota- 
tion that  a  line  OP  makes  in  turning  about  a  point  0  from 
an  initial  position  OX  until  it  comes  to  rest  in  a  terminal 
position  OQ  is  said  to  be  the  angle  described  (or  generated) 
by  the  line  OP. 

Angles  unlimited  in  magnitude.  Since  the  revolving  line 
may  make  any  number  of  complete  revolutions  before  com- 


Fig.  30. 


Fig.  31. 


Fig.  32. 


ing  to  rest,  angles  can  be  of  any  magnitude  and  can  be 
unlimited  in  magnitude. 

When  necessary  the  number  of  revolutions  can  be  indi- 
cated as  in  Fig.  32,  which  represents  the  angle  60°  and  the 

36 


ANGLES  AND   THEIR  TRIGONOMETRIC  FUNCTIONS      37 

angle  3.360° +  60°.     The  lines  in  this  figure  may  represent 
any  angle  ft -360° +  60°,  n  denoting  any  whole  number. 

Ex.  Taking  the  initial  line  in  the  same  position  as  OX  in 
Fig.  30  draw  the  terminal  lines  of  the  angles  40°,  160°,  220°, 
325°,  437°,  860°,  1020°,  180°,  360°,  720°. 

Positive  and  negative  angles.  When  the  turning  line 
revolves  in  the  counterclockwise  direction  (as  in  Fig.  30), 
the  angle  described  is  called  positive  and  is  given  the  plus 
sign;  when  the  turning  line  revolves  in  the  clockwise  direc- 
tion (as  in  Fig.  31),  the  angle  described  is  called  negative 
and  is  given  the  minus  sign. 

Ex.  Taking  the  initial  line  in  the  same  position  as  OX  in 
Fig.  30  draw  the  terminal  lines  of  the  angles  140°,  —200°, 
-430°,  335°,  -850°,  820°. 

Coterminal  angles.  Angles  having  the  same  initial  and 
terminal  lines  respectively  are  called  coterminal  angles. 

Ex.  Show  that,  if  the  same  initial  line  be  taken,  the  angles 
40°,  -320°,  400°,  760°,  -1040°  are  coterminal. 

Congruent  angles.  Angles  differing  by  multiples  of  360° 
are  called  congruent  angles. 

Ex.  Show  that  the  angles  in  the  preceding  exercise  are 
congruent. 

Complementary  angles.  Supplementary  angles.  Angles 
whose  sum  is  90°  are  called  complementary  angles;  angles 
whose  sum  is  180°  are  called  supplementary.     In  other  words, 

the  complement  of  angle  A  =  90°—  A ; 
the  supplement  of  angle  ^4  =  180°— A 

Angles  in  the  various  quadrants.     In  Fig.  33,  XXf  and 


BNs 

Y 

>PX 

/ 

V 

X 

7 

\« 

p.     (,>* 

38  ELEMENTS  OF  PLANE  TRIGONOMETRY 

YYf  are  at  right  angles  and  OX  is  the  initial  line.  In  this 
figure  XOY,  YOX',  X'OY',  Y'OX, 
are  called  the  first,  second,  third, 
and  fourth  quadrants,  respectively. 
When  the  turning  line  ceases  its  x'- 
revolution  at  some  position  be- 
tween OX  and  OY,  the  angle 
described  is  said  to  be  an  angle 
in  the  first  guadrant;  when  the 
final  position  of  the  turning  line  is 

between  OY  and  OX'  the  angle  described  is  said  to  be  in 
the  second  quadrant)  and  so  on  for  the  third  and  fourth 
quadrants. 

EXAMPLES 

1.  Lay  off  the  following  angles:  In  the  case  of  each  angle 
name  the  least  positive  angle  that  has  the  same  terminal  line. 
Name  the  quadrants  in  which  the  angles  are  situated.  In  the 
case  of  each  angle  name  the  four  smallest  positive  angles  that 
have  the  same  terminal  line. 

(a)   137°,  785°,  321°,  930°,  840°,  1060°,  1720°,  543°,  3657°. 

(6)    -240°,  -337°,  -967°,  -830°,  -750°,  -1050°,  -7283°. 

2.  What  are  the  complements  and  supplements  of  40°, 
227°,  -40°? 

complement  of    40°=   90°-  40°=         50°; 

supplement  of    40°=  180°—  40°=     140°. 
complement  of  227°=   90° -227°=  -137°; 

supplement  of  227°=  180°-227°=  -  47°. 
complement  of  -  40°  =   90°  -  ( -  40°)  - 130° ; 
supplement  of  -40°- 180°- (-40°)  =220°. 

3.  What  are  the  complements  of  -230°,  150°,  -40°,  340°, 
75°,  83°,  12°,  -295°,  -324°;  200°,  240°,  -110°,  -167°? 

4.  What  are  the  supplements  of  the  angles  in  Ex.  3? 


ANGLES  AND   THEIR  TRIGONOMETRIC  FUNCTIONS      39 

17.  Measurement  of  angles.  There  are  three  systems  of 
angular  measure,  the  sexagesimal,  the  centesimal,  and  the 
circular. 

In  the  sexagesimal  system  the  unit  angle  is  one-ninetieth 
of  a  right  angle,  called  a  degree,  and  the  subdivisions  are 
minutes  and  seconds,  as  described  in  Art.  2.  This  measure 
is  used  in  the  solution  of  triangles  and  in  elementary  mathe- 
matics generally. 

In  the  centesimal  system  the  unit  angle  is  one-hundredth 
of  a  right  angle,  called  a  grade.  The  table  of  centesimal 
measure  is 

1  right  angle  =  100  grades. 

1  grade  =  100  minutes. 

1  minute  =  100  seconds. 

This  system,  which  is  used  in  France,  has  not  come  into 
general  use. 

In  the  circular  system  the  unit  angle  is  the  angle  which 
is  subtended  at  the  centre  of  a  circle  by  an  arc  equal  in  length 
to  the  radius.     This  angle  is  called  a  radian.     The  radian 


Fig.  34.  Fig.  35. 


measure  (often  called  the  circular  measure)  of  an  angle  is  the 
number  of  radians  it  contains.  Radian  measure  is  the 
measure  used  in  higher  mathematics.  It  is  also  used  in 
various  problems  and  discussions  in  elementary  mathematics. 


40  ELEMENTS   OF  PLANE  TRIGONOMETRY 

18.  Value  of  a  radian.  Relation  between  radian  measure 
and  degree  measure.  In  Fig.  35  arc  AB  =  radius,  and  thus 
angle  AOB  =  a  radian.     By  geometry, 

angle  AOB  arc  A B 

complete  angle  about  0     length  of  circle' 

a  radian        radius         1 
Le*'  360°    =  2ttX  radius  =  2k' 

180° 
/.     a  radian  = =57°  17'  44".8.  (1) 

IT 

Thus  a  radian  has  a  constant  value,  and  accordingly  can 
be  used  as  a  unit.      From  (1) 

ir  radians  =  180°.  (2) 

Also,  from  (1)  or  (2), 

l°  =  ^r  radians. 

By  means  of  relation  (2)  an  angle  expressed  in  the  one 
measure  can  be  expressed  in  the  other. 

Notation.  An  angle  2  radians  is  expressed  2r  or  2C  (r 
from  radian,  c  from  circular).  The  symbol  r,  or  c,  is  often 
omitted.    For  instance,  angle  n  denotes  the  angle  n  radians 

i.e.,  180°;  angle  «r  denotes  the    angle  ~   radians,  i.e.,  90°; 

angle  6  denotes' an  angle  containing  6  radians. 

Radian  measure  as  ratio  of  subtended  arc  to  radius.  Length 
of  arc.  Since  the  arc  of  a  circle  equal  to  the  radius  subtends 
at  the  centre  an  angle  equal  to  a  radian,  the  number  of  times 
an  arc  contains  the  radius  is  the  radian  measure  of  its  subtended 
angle;  i.e., 

_  ■„  .  .      length  of  subtended  arc 

Number  of  radians  in  angle  =  - — : — — 3 — -r. .     (3) 

length  of  radius  v 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS      41 


If  a  =  length  of  arc,  r  =  length  of  radius,  and  0= radian 
measure  of  the  angle,  (3)  may  be  written 


e  = 


a 


v 
a=rQ, 


(4) 

(5) 


From  this 

In  words,    arc  =  radius  X  radian  measure  of  angle. 

Formula   (4)   derived  otherwise.     In  Fig.   36  AOB  is  a 
radian.      Let   r  =  length   of   radius,    a  =  length   of   arc    AP, 

6  =  radian  measure  of  angle  AOP. 
By  geometry, 

angle  AOP  _  arc  AP 
angle  A  OB  ~  arc  AB} 

angle  AOP    a 
1  radian  ~ r' 

number  of  radians  in  AOP,  0=2. 


i.e., 


Fig.  36. 


o-7r,  —  107T,  in 
o 


EXAMPLES 

1.  How  many  degrees  are  there  in  2.5  radians? 

2.  How  many  radians  are  there  in  231°? 

3.  Express  J*",  4r,  Jr  in  degrees. 

t-i  x-l  *        X     7C     X     K     2    ' 

4.  Express  the  angles  -,  -,  j,  -,  ^rr,  3tt, 
degrees. 

5.  Express  in  radians  83°  20',  142°  30'. 

6.  Express  in  radians  (in  terms  of  n)  45°,  210°,  300°,  120°, 
225°. 

7.  What  is  the  radian  measure  of  the  angle  which  at  the 
centre  of  a  circle  of  radius  1J  yds.  subtends  an  arc  of  8  in.? 
Also  express  the  angle  in  degrees. 


42  ELEMENTS   OF   PLANE  TRIGONOMETRY 

8.  Find  the  radius  of  the  circle  in  which  an  arc  12  in. 
subtends  an  angle  2  radians  at  the  centre. 

9.  The  radius  of  a  circle  is  10  in.  How  long  is  the  arc  sub- 
tended by  an  angle  of  4  radians  at  the  centre? 

10.  How  long  does  it  take  the  minute  hand  of  a  clock  to 
turn  through  —  If  radians? 

18a.  Motion  of  a  particle  in  a  circle. 

Let  r  ft.*  =  radius  of  the  circle  on  which  the  particle  is 

moving; 
s  ft.  =  length  of  arc  traversed ;  and 
6  radians  =  angle  swept  over  by  the  radius  joining  the 

moving  point  to  the  centre  of  the  circle. 

Then,  by  Art.  18,  Eq.  *(5), 

s  =  rQ. 

It  should  be  noted  that  this  equation  is  not  true  unless 
the  radius  and  the  arc  are  measured  in  the  same  unit  of 
length  and  the  angle  is  measured  in  radians. 

Again,  suppose  the  particle  is  moving  at  a  uniform  rate 
on  a  circle  of  radius  r  ft.* 

Let     v  ft.  per  sec.  =  the  linear  velocity  of  the  particle  in 
its  path, 
and  (o  radians  per  sec.  =  its  angular  velocity. f 

Then  v=ro>; 


and  thus  «  =  ^» 

*  Or  whatever  may  be  the  unit  of  length. 

f  The  angular  velocity  of  the  particle  is  the  number  of  radians  swept 
over  in  a  unit  of  time — in  this  instance  a  second — by  the  radius  joining 
the  moving  point  to  the  centre  of  the  circle. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS      43 

Suppose  the  angular  velocity =N  revolutions  per  minute. 

Since  1  revolution  =  2n  radians, 

N  revolutions  per  minute  =  licN  radians  per  min. 

irJV      ,. 
=-sr  radians  per  sec. 


EXAMPLES 

1.  A  wheel  of  a  carriage  which  is  travelling  at  the  rate 
of  8  miles  per  hour  is  3  ft.  in  diameter.  Find  the  angular 
velocity  of  any  point  of  the  wheel  about  the  axle,  in  radians 
per  second. 

2.  Compare  the  angular  velocities  of  the  hour,  minute,  and 
second  hands  of  a  watch. 

3.  Express  in  terms  of  radians  per  minute  an  angular 
velocity  of  25°  per  second. 

4.  A  wheel  makes  300  revolutions  per  hour.  Express  its 
angular  velocity  (a)  in  radians  per  sec. ;   (6)  in  degrees  per  min. 

5.  An  automobile  having  a  28-in.  wheel  travels  at  a  speed 
of  15  miles  an  hour.  What  is  the  angular  speed  of  the  wheel 
about  the  axle  (a)  in  revolutions  per  minute?  (6)  in  radians  per 
second? 

6.  A  belt  travels  over  a  30-in.  pulley  which  is  carried  on  a 
shaft  making  300  revolutions  per  minute.  Find  the  linear 
speed  of  the  belt  in  feet  per  second. 

19.  Convention  for  signs  of  lines  in  a  plane.  In  Fig.  37 
X'OX  is  a  horizontal  line  to  which  YOY'  is  drawn  at  right 


44 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


Pi 


angles.  Lines  measured  horizontally  towards  the  right  are 
taken  as  positive,  and  accordingly  lines 
measured  horizontally  towards  the  left 
are  negative.  Thus  OM1}  OM±,  M20 
are  positive,  OM2,  OM3,  MxO  are 
negative.  Lines  measured  vertically 
upward,  as  MXP\,  M2P2,  are  taken  as 
positive,  and  lines  measured  verti- 
cally downward,  as  M3PS,  M±P4,  are 
negative. 


M, 


X    M* 


P3 


O  M1 


MtX 


Y' 
Fig.  37. 


20.  Trigonometric  functions  defined  for  angles  in  general. 

In  Figs.  38-41  the  horizontal  line  OX  is  taken  as  the  initial 
line,  and  angles  in  the  first,  second,  third,  and  fourth  quad- 

Fig.  38.  Fig.  39. 

ft  <u 


7^ 
/    1 

X'  M   - 

O    +    M    X 
.     0               X 

X'  M    ~    O 


M  X 


Fig.  40. 


Fig.  41. 


rants  respectively  are  described.  In  each  angle  any  point 
P  in  the  terminal  line  OQ  is  taken,  and  MP  is  drawn  at 
right  angles  to  OX.    Thus  (Art.  10)  in  each  figure 

OM  is  the  horizontal  projection  of  OP, 

MP  is  the  vertical  projection  of  OP. 

The  lines  OM,  MP  receive  signs  in  accordance  with  the 
convention  in  Art.  19  and  OP  is  taken  as  positive.  For  each 
figure,  the  angle  XOQ  being  denoted  as  A,  the  six  ratios 


ANGLES  AND  THEIR  TRIGONOMETRIC    FUNCTIONS      45 

formed  by  means  of  the  lines  OM,  MP,  OP  are  called  trigo- 
nometric functions  of  A,  as  follows: 

the  sine  of  angle  A  =  sin  A  =  -^-p  > 

the  cosine       of  angle  A  =cos  -^  =  ~<jp» 

MP 

the  tangent     of  angle  J.  =  tan  .4  =  Q]j£i 

the  cotangent  of  angle  A  =  cot  ^4.  =  ]^p  > 

OP 

the  secant       of  angle  A  =  sec  A  =  q^  > 

the  cosecant    of  angle  A  =  esc  ^4.  =  jjbfp  • 

Two  other  trigonometric  functions  are  occasionally  used, 
viz.: 

the  versed  sine  of  angle  A  =  vers   A  =  1  —  cos  A, 
the  cover sed  sine  of  angle  A  =  covers  A  =  l  —  sin  A. 

Inspection  of  Figs.  38-41,  in  connection  with  the  defini- 
tions, shows  that  the  trigonometric  functions  of  angles  in 
the  various  quadrants  have  the  signs  stated  in  the  following 
table : 


Quadrant. 

sin  A 

cos  A 

tan  A 

cot  A 

sec  A 

cosec  A 

I. 

+ 

+ 

+ 

+ 

+ 

+ 

II. 

+ 

- 

- 

- 

- 

+ 

III. 

- 

- 

+. 

+ 

- 

- 

IV. 

- 

+ 

- 

- 

j_ 

- 

46 


ELEMENTS   OF  PLANE  TRIGONOMETRY 


EXAMPLES 


1.  Tell  the  signs  of  the  following  functions: 

(a)  sin  100°;         (b)  cos  220°;  (c)  tan  230°; 

(d)  sec  340°;         (e)  cot  (-130°);         (/)   esc  560°. 

2.  Describe  the  angle  480°,  and  find  the  values  of  its  sine, 
cosine,  and  tangent. 

3.  Describe  the  angle  945°;  and  find  the  values  of  its  sine, 
cosine,  and  tangent. 

4.  In  which  quadrants  may  an  angle  lie,  if : 

(a)  its  sine  is  positive;  (b)  its  cosine  is  negative; 

(c)   its  tangnet  is  positive;     (d)  its  tangent  is  negative. 

5.  In  what  quadrant  must  an  angle  lie,  if: 

(a)  its  sine  and  cosine  are  negative; 

(b)  its  sine  is  positive  and  tangent  negative; 

(c)  its  sine  is  negative  and  tangent  positive; 

(d)  its  cosine  is  positive  and  tangent  negative. 

6.  Construct  angle  A,  when  sin  A  =  f .  Find  the  remaining 
functions  of  A.  The  definition  of  the  sine,  Art.  20,  shows  that 
for  this  angle 

M?=+3,         OP=4. 

The  construction  then  is  as  indi- 
cated in  Fig.  42.  There  are  thus 
two  sets  of  angles  whose  sines  are  J, 
viz.,  the  angles  whose  terminal  line 
is  OP,  and  the  angles  whose  terminal 
line  is  OPi. 

Each  set  of  angles  is  unlimited  in 
number,  and  the  angles  in  it  differ  from  one  another  by  mul- 
tiples of  360°. 


Fig.  42. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS       47 
For  an  angle  having  OP  for  terminal  line, 


cos  A  = 


Vf 


tan  A 


sec  A  =  - 


vf 


csc  A 


cot  A  = 


VI 


vT     ~~    3 

For  an  angle  having  OPi  for  terminal  line, 
Vf        .  3 


cos  A=  — 


tan  A  = 


VT 


cot  A  =  — 


Vf 

3  ' 


Xl  M 


4  A        4 

secA= =,       cscA  =  — . 

Vf  3 

3      r  3    —3     3    1 

7.  Construct  angle  A,  given  tan  A=— —.      Note  — j="4_==34- 

The  definition  of  the  tangent,  Art.  20,  shows  that  for  this  angle, 

MP=-3     and     0M=    *4, 

or 

MP=3       and       0Af=-4. 

The  construction  then  is  as 
indicated  in  Fig.  43.  There 
are  thus  two  sets  of  angles 
whose  tangents  are  — },  viz., 
the  angles  whose  terminal  line  is  OP  and  the  angles  whose 
terminal  line  is  OPx.     The  angles  in  each  set  are  congruent. 

8.  Calculate  the  remaining  functions  of  the  angles  in  Ex.  7. 

9.  Construct  the  angles  which  have  the  following  functions, 
and  calculate  the  remaining  functions: 

(a)  cos  x=  - s-;     (&)  sin#=—  -p-;     (c)  cos#=-r-; 


Fig.  43. 


48 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


(d)  sec  x=p 


(.)  tan,=i.        [Note.  J-±j-=£] 


10.  Construct  the  angles  and  find  their  remaining  functions 
in  the  following  cases: 

3  2 

(a)  sin  x  =  =■  and  cos  x  negative ;  (jb)  cos  x = ^  and  sin  a;  negative ; 

3 

(c)  tan  x = ^-  and  sin  #  negative ;  ((f)  sec  a; = 2  and  tan  x  negative. 

21.  Line  definitions  of  the  trigonometric  functions.  Geo- 
metrical representation  of  the  functions.  In  Fig.  44  XOP 
is  an  angle  of  x  radians.  A  circle, 
whose  radius  is  a  unit  in  length, 
is  described  about  0.  This  circle 
is  called  a  unit  circle.  Angle 
XOF  =  90°.  From  A  and  B  tan- 
gents AT,  BS  are  drawn  to  meet 
the  .terminal  line  OP.  From  P, 
the  intersection  of  the  terminal 
line  with  the  circle,  PM  is  drawn 
at  right  angles  to  OX.    Then 


7 

B 

cot  X 

SA 

t 

N 

sseex 

I 

S 
8 

C 

cosX  M 

A.X 

Angle  in  First  Quadrant. 

Fig.  44. 


sm   a>  = 


cos  cc  = 


tan  »i 


MP 
OP(-l) 

Oflf 

OP(-l) 

'Oi(-l) 


=  MFy 


=  OMy 


=AT, 


cot  ^-Mp-0jB(  =  1)-^» 


*  Since  the  triangles  OMP,  OBS,  are  similar, 
f  Thus  also,        vers  x  =  l  —  cos  £  =  OA  —  OM  —  M A ; 
covers  x  =  1  —  sin  x  =  OB  —  MP  =  NB. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS      49 
OT 

sec  *=oXC=T)       =ot' 

OP        OS* 

csc  —Ep-OTj^-oi*. 

. ,  arc  AP 

Also  ^""n A  ( —  n=arc  AP* 

Since  the  radius  is  of  unit  length  the  lengths  of  these 
lines  are  expressed  in  the  same  numbers  as  the  ratios  which 
are  the  trigonometric  functions.  Accordingly  the  lines  can 
represent  the  functions. 

Also,  on  the  same  scale,  the  arc  represents  the  radian 
measure  of  the  angle.     (Compare  definition  (3),  Art.  18.) 

The  line  definitions  of  the  trigonometric  functions  of  an 
angle  may  be  thus  expressed,  the  angle  being  at  the  centre 
of  a  unit  circle : 

The  sine  is  the  length  of  the  perpendicular  drawn  from  the 
extremity  of  the  terminal  radius  to  the  initial  radius. 

The  cosine  is  the  length  of  the  line  from  the  centre  of  the 
circle  to  the  foot  of  this  perpendicular. 

The  tangent  is  the  length  of  the  tangent  drawn  at  the  extrem- 
ity of  the  initial  radius  from  this  extremity  to  where  the  tangent 
meets  the  terminal  radius  produced. 

The  secant  is  the  length  from  the  centre  along  the  terminal 
line  to  where  it  meets  the  tangent  drawn  at  the  extremity .  of  the 
initial  radius. 

The  cotangent  is  the  length  of  the  tangent  drawn  at  the 
extremity  of  the  radius  which  makes  an  angle  +  90°  with  the 
initial  radius,  from  this  extremity  to  where  this  tangent  meets 
the  terminal  radius  produced. 

The  cosecant  is  the  length  from  the  centre,  along  the  terminal 
line  to  where  it  meets  the  tangent  drawn  at  the  extremity  of  the 
radius  which  makes  an  angle  +90°  with  the  initial  radius. 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


Figs.  45,  46,  47  show  the  lines  which  represent  the  trigo- 
nometric functions  of  angles  in  the  second,  third,  and  fourth 
quadrants  respectively.  The  lines  are  named  as  in  Fig.  44. 
The  line  representing  the  tangent  is  always  drawn  from  A, 


\S  cotx  B 

Y 

~\cosxM 

A 

3 

1 

M 
/t 

\-2 

Angle  in  Second  Quadrant. 

Fig.  45. 


B 

cot  X     S/J 

'T 

mf     R- 

| 

1   M       fi 

C08X)>S 

0           ]ax 

\l 

-^^_ 

Angle  in  Third  Quadrant. 

Fig 

.46. 

and  the  line  representing  the  cotangent  is  always  drawn 
from  B,  to  the  terminal  line  of  the  angle.  The  signs  of  OM, 
MP,  AT,  BS  follow   the  convention  in  Art.  19,   and   the 


*  Angle  in  Fourth  Quadrant. 

Fig.  47.  Fig.  48. 

direction  from  0  towards  P  is  taken  as  positive.  Inspection 
of  Figs.  44-47  shows  that  the  functions  of  angles  in  the, 
various  quadrants  have  the  signs  set  down  in  the  table  ♦in 
Art.  20  * 

*  The  line  definitions  of  the  trigonometric  functions  were  employed 
before  the  ratio  definitions  were  suggested. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS       51 

22.  Changes  in  the  values  of  the  functions  when  the  angle 
varies.  Limiting  values  of  the  functions.  In  Fig.  48  AOP 
is  a  variable  angle  x,  and  a  unit  circle  is  described  about  0. 

MP 
The  sine.     Now      sin  x  =  7jpTZT\  =  M?- 

When  angle  x  approaches  0°  as  a  limit,  MP  passes  through 
the  values  M2P2,  M1P1  to  zero  as  a  limit.     Thus 

sin  0°  =  0. 

When  OP  revolves  positively  from  OA  toward  OB  as  a 
limit,  x  increases  from  0°  to  90°  as  a  limit,  and  MP  increases 
from  zero  through  MxP1}  M2P2,  M^P^  to  OB(  =  l)  as  a  limit. 
Thus 

sin  90°  =  1. 

When  OP  revolves  from  OB  to  OA  lf  x  increases  from  90° 
to  180°,  and  MP  decreases  from  0£(  =  1)  through  M&PB, 
MqP6  to  zero.     Thus 

sin  180°  =  0. 

When  OP  revolves  from  OAi  to  OBi,  x  increases  from 
180°  to  270°  and  MP  changes  from  zero  through  M7P7, 
M8P8  to  OJBi(-  -1),  and  thus 

sin  270°=  -1. 

When  OP  revolves  from  OBi  to  OA,  x  varies  from  270° 
to  360°,  and  MP  changes  from  0#i(=-l)  through  M9P9, 
^10^10  to  zero.     Thus 

sin  360°  =  0. 

The  cosine.     Now 

0M        rw 

cos  x  —  A)p/_-i\  =  OM. 


52  ELEMENTS  OF  PLANE  TRIGONOMETRY 

When  angle  x  approaches  0°  as  a  limit,  OM  passes  through 
the  values  OM2,  OMx  to  OA  (  =  1)  as  a  limit.     Thus 

cos  0°  =  1. 

When  OP  revolves  positively  from  OA  to  OB,  x  increases 
from  0°  to  90°  and  OM  decreases  from  OA(  =  l)  through 
OMh  OM2,  OMs,  OM4,  to  zero.    Thus 

cos  90°  =  0. 

When  OP  revolves  from  OB  to  OA  u  x  increases  from  90° 
to  180°  and  OM  changes  from  zero  through  OM5,  OM6  to 
OA^-1).    Thus 

cos  180°-  -1. 

When  OP  revolves  from  OA  i  to  OB1}  x  increases  from 
180°  to  270°  and  OM  changes  from  OAi(  =  -l)  through 
OM7,  OM8  to  zero.     Thus 

cos  270°  =  0. 

When  OP  revolves  from  OBx  to  OA}  x  increases  from  270° 
to  360°  and  OM  changes  from  zero  through  OM9}  OMXQj 
toOA  =  l.    Thus 

cos  360°  =  1. 

The  tangent.  First  method,  using  the  ratio  definitions. 
Now 

MP 
tan*=OM' 

When  angle  x  approaches  0°  as  a  limit,  MP  approaches 
zero  and  OM  approaches  0 A  ( =  1 ) .    Thus 

tan0°  =  0. 


ANGLES   AND  THEIR  TRIGONOMETRIC  FUNCTIONS      53 

When  x  increases  from  0°  to  90°,  MP  is  approaching 

MP 

OB=(l)  and  OM  is  approaching  zero.     Thus  jyjur,  which  is 

positive  during  this  change,   is  constantly  increasing  and 
becomes  unlimited  in  value.     Thus 


tan  90°  =  oo. 

When  x  is  increasing  from  90°  to  180°,  MP  is  changing 
from  OJ5(  =  l)   to  zero  and  OM  is  changing  from  zero  to 

MP 

OA  i  ( =  —  1 ) .     Thus  Y^Tj,  which  is  negative  during  this  change, 

changes  from  —  oo  to  0.    Thus 

tan  180°  =  0. 

N.B.  When  x  is  increasing  from  0°  to  180°,  tanz=  +oo 
when  x  reaches  90°,  and  tan  x  =  —  oo  when  x  leaves  90°. 
Thus  tan  x  changes  sign  when  x  is  passing  through  the  value 
90°.  If  x  is  decreasing  from  180°  to  0°,  tan  x  =  —  oo  when 
x  reaches  90°,  and  tanz=+oo  when  x  leaves  90°.  Thus 
the  sign  of  tan  x  when  z  =  90°,  depends  on  the  way  in  which 
x  has  approached  the  value  90°. 

When  x  is  increasing  from  180°  to  270°,  MP  is  changing 
from  zero  to  OBi   ind  OM  is  changing  from  OAi  to  zero. 

MP 

Thus  :~  T7,  which  is  positive  during  this  change,  changes  from 

0  to  an  unlimited  value.     So 

tan  270°  =  oo . 

When  x  is  increasing  from  270°  to  360°,  MP  is  changing 
from  OB i  to  zero,  and  OM  is  changing  from  zero  to  OA. 


54 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


MP 

Thus  :~  T7,  which  is  negative  during  this  change,  changes  from 

—  oo  to  0.     So 

tan  360° =0. 

Tan  270°,  like  tan  90°,  has  an  unlimited  value  whose  sign 
is  positive  or  negative,  according 
to  the  way  in  which  the  angle 
has  approached  the  value  270°.  f 
The  tangent.  Second  method, 
using  the  line  definitions.  In  Fig. 
49,  the  variable  angle  AOP  =  x. 
A  unit  circle  is  drawn  and  the 
tangent  drawn  at  A.    Now 


tanz 


AT 


Fig.  49. 


OA(-l) 


=  AT. 


When    x    approaches   0°,   AT 
passes  through  AT2y  AT\  to  zero;  and  thus 

tan  0°=0. 

When  x  changes  from  0°  to  90°,  AT  passes  through  ATi, 
AT 2,  ATzj  AT 4,  A T5  to  an  unlimited  value;  and  thus 

tan  90°  =  oo. 


The  student  can  trace  by  means  of  Figs.  45,  46,  47  the 
further  changes  in  tan  x  when  x  increases  from  90°  to  360°. 

The  tracing  of  the  changes  in  cot  x,  sec  x,  esc  x,  as  x 
changes  from  0°  to  360°,  is  left  as  an  exercise  to  the  student. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS       55 

The  changes  for  the  above  six  functions  when  the  angle  is 
increasing  from  0°  to  360°  are  indicated  in  the  following  table: 


In  the  second  quadrant  the 

Y 

In  the  first  quadrant  the 

sine           decreases  from 

lto 

0 

sine            increases  from 

Oto 

1 

cosine        decreases  from 

Oto- 

-1 

cosine        decreases  from 

lto 

0 

tangent     increases  from- 

-oo  to 

0 

tangent     increases  from 

Oto 

00 

cotangent  decreases  from 

Oto- 

-  oo 

cotangent  decreases  from 

oo  to 

0 

secant        increases  from- 

-ooto- 

-  1 

secant        increases  from 

lto 

00 

cosecant    increases  from 

lto 

00 

•  cosecant    decreases  from 

oo  to 

1 

y 

V 

A\ 

0 

A 

In  the  third  quadrant  the 

In  the  fourth  quadrant  the 

sine            decreases  from 

Oto- 

- 1 

sine            increases  from- 

-  lto 

0 

cosine        increases  from- 

-  lto 

0 

cosine        increases  from 

Oto 

1 

tangent     increases  from 

Oto 

00 

tangent     increases  from- 

-oo  to 

0 

cotangent  decreases  from 

qo  to 

0 

cotangent  decreases  from 

Oto  - 

-00 

secant       decreases  from  - 

-  lto- 

■  oo 

secant        decreases  from 

oo  to 

1 

cosecant    increases  from- 

■  ooto- 

-  1 

cosecant    decreases  from - 

-  lto  - 

-00 

Yx 


23.  Periodicity  of  the  trigonometric  functions.  When 
the  angles  360°  to  720°  are  described  the  sine  goes  through 
all  its  changes  in  the  same  order  as  when  the  angles  0°  to 
360°  are  described.  Also  the  sine  repeats  the  same  value 
every  time  the  angle  changes 
by  360°,  i.e.,  2£  That  is,  n 
denoting  a  whole  number 

sin  (n  •  360° + x)  =  sin  x, 


i.e. 


sin  (2mr +x)      =  sin  x. 
Accordingly  the  sine  is  said 


Fig.  50. 


to  be  a  periodic  function,  and  its  period  is  2x. 


56 


ELEMENTS  OF   PLANE  TRIGONOMETRY 


Similarly,  the  cosine,  secant,  cosecant  have  the  period  2n. 

When  the  angles  180°  to  360°  are  described  the  tangent 
goes  through  all  its  changes  in  the  same  order  as  when  the 
angles  0°  to  180°  are  described.  Also  the  tangent  repeats 
the  same  values  every  time  the  angle  changes  by  180°,  i.e., 
it.    That  is,  n  denoting  any  whole  number, 


i.e., 


tan  (n  •  180° + x)  =  tan  x, 
tan  {nn  +  x)       =  tan  x. 


Accordingly  the  tangent  is  a  periodic  function,  and  is 
period  is  it.    Similarly,  the  cotangent  has  the  period  n. 

24.  Graphs  of  the  functions. 

Graph  of  sin  x.     From  the  equation, 

y  =  smx 

form  corresponding  pairs  of  x  and  y.    Convenient  values  to 

TZ    TZ    TZ    tz    2n 


take  for  x  are  30°,  45°,  60°,  90°,  120°, . 
Thus: 


.  ,i.e. 


'6'4;3'  2'   3 


X 

0 

it 

"6 

Tt 

T 

It 

it 
~2 

27t 

3 

3tt 
4 

5tt 
6 

it 

7 
6* 

3 
2* 

7 
I* 

2tt 

9ar 
4 

Tt 

y 

0 

.5 

.71 

.87 

1 

.87 

.71 

.5 

0 

-.5 

-1 

-.71 

0 

.71 

-.71 

Choose  any  convenient  length  to  represent  the  angle  2n 
(360°) ;  *    mark  the  points  corresponding  to    angles  ^,   jt 

TZ 

^,  .  .  .  ;    at  these  points  erect  ordinates  representing  the 
o 

*  E.g.,  one  such  length  that  may  be    taken  is  the  length  of  a  unit 
circle. 


ANGLES  AND  THEIR   TRIGONOMETRIC  FUNCTIONS      57 


corresponding  values  of  y;  draw  a  smooth  curve  through  the 
ends  of  these  ordinates. 

This  curve,  Fig.  51,  is  called  "The  curve  of  sines11  or 
"the  sine  curve.11 

A  very  convenient  way  to  draw  the  ordinates  corre- 
sponding to  values  of  x  is  to  take  the  sines  from  the  unit 
circle  (Art.  21),  as  indicated  in  Fig.  52. 

Graph  of  cos  x.     From  the  equation 
2/  =  cos  x 
form  corresponding  values  of  x  and  y,  and  proceed  as  in  the 
case  of  the  sine.     The  ordinates  can  also  be  obtained  readily 
by  using  a  unit  circle.     The  cosine  curve  is  in  Fig.  53. 
Y 


Fig.  51. — Graph  of  sin  x. 

rr 


Fig.  53. — Graph  of  cos  x. 

Graph  of  tan  oc.     From  the  equation 
y  =  tan  x 


58  ELEMENTS  OF  PLANE  TRIGONOMETRY 

form  corresponding  values  of  x  and  y;  thus 


X 

0 

30° 

Tt 

"6 

45° 

St 

4 

60° 
Tt 
~3~ 

90° 

it 
~2 

120° 
2 

3* 

135° 
3 

4* 

180° 

Tt 

210° 
7 
6* 

270° 

~2 

300° 
5 

360° 

2it 

y 

0 

.58 

1 

1.73 

CO 

-1.73 

-1 

0 

.58 

00 

-1.73 

0 

Then,  on  proceeding  as  in  the  case  of  the  sine  the  graph 
in  Fig.  54  is  obtained. 

Y 


Fig.  54. — Graph  of  tan  x. 

By  using  a  unit  circle,  see  Fig.  49,  the  ordinates  can  be 
drawn  quickly. 

Graphs  of  cot  as,  sec  acy  cosec  oc.  These  can  be  obtained 
by  proceeding  as  in  the  case  of  the  preceding  graphs.  They 
are  shown  in  Figs.  55,  56,  57. 


Fig.  55. — Graoh  of  cot  x. 


ANGLES  AND  THEIR  TRIGONOMETRIC   FUNCTIONS       59 
F 


1  I 

Fig.  56. — Graph  of  sec  x. 


Fig.  57. — Graph  of  esc  x. 

EXAMPLES 

1.  Draw  various  graphs  for  sin  6,  cos  6,  tan  0,  by  varying 
the  scales  used  in  representing  radians  and  trigonometric 
functions. 

2.  Draw  graphs  of  the  following: 

(a)  sin#  +  cos#;      (b)  sin  6— cos  6)      (c)  sin  26;       (d)  cos  30; 

0)  sin  -;     (/)  tan  2« 


GO  ELEMENTS  OF   PLANE  TRIGONOMETRY 

3.  Draw,  with  the  same  axes  of  reference,  graphs  of  sin  6 
and  cos  6;  and  from  your  figure  obtain  values  of  6  between  0° 
and  360°  for  which  (1)  sin  0=cos  0;  (2)  sin  6+ cos  6  =  0. 

Also  with  the  help  of  this  figure  draw  the  graph  of 
sin  0-j-cos  6. 

25.  Relations  between  the  trigonometric  functions  of  an 
angle.  The  relations  between  the  trigonometric  functions  of 
an  acute  angle  were  set  forth  in  Art.  7.  These  relations  hold 
for  the  ratios  of  any  angle. 

A.  Inspection  of  the  definitions,  Art.  20,  shows  the 
reciprocal  relations,  namely: 

tan^cot^i  =  l;     cos  A  sec  -4  —  1;     sin  A  esc  A  —  1,     (1) 

*    A  1  1  A  1 

i.e.,  cot  A  =.- t:  secA  = 7;  esc  A=- — r. 

tan  A'  cos  A'  sin  A 


B.  In  each  of  the  figures  in  Art.  20, 

MP                                     OM 

MP     OP     smA                 OM     OP      cos  A 
t&nA     OM~OM~cosA>  cotA~MP~MP~smA- 

(2) 

OP                                      OP 

C.  In  each  of  the  figures  in  Art.  20, 
MP2  +  OM2  =  OP2. 


On  dividing  both  members  of  this  equation  by,   OP 

. 2     ■ 2 

OM  ,  MP  ,  in  turn,  and  following  the  same  process  as  that 
adopted  in  Art.  7,  it  results  that 


sin2  A  +  cos2  A  =  l] 

sec2. 1  =  1  + tan2  .4; 
cosec2  ^4=1  +  cot2  A. 


(3) 


Relations  C  also  follow  directly  from  the  line  definitions, 
Art.  21,  as  shown  in  Fig.  44  or  Figs.  45-47. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS      61 

,   EXAMPLES 

1.  Given  that  sinA  =  f;  find  the  other  functions  of  A  by 
means  of  the  relations  shown  in  this  article. 

[In  Ex.  6,  Art.  20,  this  problem  is  solved  geometrically; 
here  it  is  solved  algebraically.] 

±V7  .         1  4 


cos 


A  =  ±  Vl  — sin2  A  =  — - — :     sec  A  = 


cos  A     -j-  \/y  ' 


1        4  .     sin  A         3 

cosecA=— — 7  =  5-;     tanA  = 


sin  A     3  '  cos  A      ±  V7  ' 

cot  A  =  - j= — — . 

tan  A  6 

Since  the  given  sine  is  positive,  the  corresponding  angles 
are  in  the   first   and   second   quadrants.     Hence   the   double 
values  of  the  calculated  ratios  may  be  paired  as  follows: 
sin  A        cos  A         tan  A       cot  A        sec  A     cosec  A 

3  +VY  3  VT  4  4 

4  4  a/7  3  v7  3 

3  -V7  3  -VT  4  4 

4  4  V7  3  Vf  3 

Find  the  other  ratios  algebraically,  and  verify  the  results 
geometrically,  when: 

2.  cos  A     =—  f.  3.  tanA=-f. 

4.  cosec  A=— 5.  5.  cotA=  —  3. 

Find  the  other  ratios  algebraically,  and  verify  the  results 
geometrically,  when  angle  A  satisfies  the  following  pairs  of 
conditions : 

6.  sin  A  =  J      and     tan  A  negative. 

7.  tan  A  =  \/3     and  sec  A  negative. 

8.  cosA=— J     and  sin  A  positive. 

9.  sin  A  =  —  f     and     tan  A  positive. 


62 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


10.  If  sin  0=    a  .  „   '    .  „  „,  prove  that 


m2  +  2mn  +  2n2' 


tan  0=  ± 


m2  +  2mn 
2mn+2n2' 


a2—b2 

11.  If  cos  6=   0  ,  rst  find  sin  0  and  tan  0. 

Verify  each  of  the  following  relations  by  reducing  the  first 
member  to  the  second: 

12.  cos  x  tan  x= sin  x. 

13.  sec  x— tan  x> sin  z= cos  x. 
cot2  A 


14. 


=  cos2  A. 


l  +  cot2A 
16.  (l  +  tan2z)  cos2  0=1. 
16.  cos4  0-sin4  0+1  =  2  cos2  0. 
1  1 


17 


+  ■ 


=  1. 


tan2  B+l  '  cot2£+l 
18.  sec4  0-tan4  6=2  sec2  0-1. 

26.  Functions  of  —  A,  90°  +  A,  180°  +  A  in  terms  of  func- 
tions of  A,  A  being  any  angle. 

Functions  of  —A.     Describe  the  angles  A,  —  A,  OP  being 

p  P-  Pn  R 


Fig.  59 


Fig.  60. 


Fig.  58. 

the  terminal  line  of  A,  and  OPi  the  terminal  line  of  —A. 
In  Figs.  58,  59,  60,  61,  A  is  in  the  first,  second,  third,  fourth 
quadrants  respectively. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS      63 


Take  OPl  =  OP,  and  draw  PM,  PXMX  at  right  angles  to 
OX.    Then  in  each  figure 


OM^OM, 


M,P, 


MP. 


Thus,  for  an  angle  A  in  any  quadrant, 

sin  (--4)  =  -gfp-=-^p=-sin^; 


cos  {—A) 
So  also    tan  (—  A) 


0Ml 


OM 


=     cos  A. 


OPi  OP 

-tan  A;    cot  (—A)  =  —cot  A; 
sec  (—A)  =  sec  A;         esc  (—A)  =  —esc  A. 


Functions  of  90°  —  ^.  Describe  the  angles  A,  90°— A, 
OP  being  the  terminal  line  of  A,  and  OP\  the  terminal  line 
of  90°-  A.  In  Figs.  62,  63,  64,  65,  A  is  in  the  first,  second, 
third,  fourth  quadrants  respectively. 


/p' 

^ 

M    Mi    0 

1 

A  * 

*  H7 

xiv 

Jf 

iy\ 

M    0^ 
Fig. 

vtf*'  V 

63.                Fig.  64. 

If,  0^>v 

Fig.  6 

i    X 

O    Mx    M  X 
Fig.  62. 

5. 

Take  OP,  =  OP,  and  draw  Pikf,  P,ilf,  at  right  angles  to 
OX.    Then  in  each  figure 

OMx=MP,  MxPx^OM. 

Thus,  for  A  in  any  quadrant,  and  so  for  all  angles  A, 

sm  (90°-A)==-^p-  =Qp=cos  A; 

/rtrtQ      „      OM,     MP     .     A 
cos  (90°-A)=-gp-=-gp  =sm  A; 


64 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


fMA     ix    MXPX    OM 
tan  (90°-A)=m^-  =]^p  =  cot  A- 

cot  (90°- A)  =w^i  =m  =  tan  i; 

sec  (90°-  A)  =-qj^=mp  =  ^  M 


esc  (90°- A) 


OiPi      OP 


=  sec  A. 


MlPl     OM 

Hence,  the  function  of  any  angle  is  the  same  as  the  co- 
function  of  its  complement. 

Functions  of  90°  +  A.  Describe  the  angles  A,  90°  +  A, 
OP  being  the  terminal  line  of  A  and  OPi  the  terminal  line 
of  90°  +  A.  In  Figs.  66,  67,  68,  69,  A  is  in  the  first,  second, 
third,  fourth  quadrants  respectively. 


Fig.  66.  Fig.  67.  Fig.  68.  Fig.  69. 

Take  OPx^OP,  and  draw  PM,  PXMX  at  right  angles  to 
OX.    Then  in  each  figure 

OMx  -  -  MP,  MxPx  =  OM. 

Hence,  for  any  angle  A7 

MiPi  _     OM 
~     OP 
MP 
~~OP  ' 

So  also,  tan  (90° + A)  -  -  cot  A ;    cot  (90° + A)  =  -  tan  A  ; 
sec  (90°  +  A)  =  -  esc  A ;    esc  (90°  +  A)=     sec  A. 


sin  (90°+ A)  =: 


OP 


cos(90°+A)=-^p- 


cos  A; 
sin  A. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS      65 


Functions  of  180° -X      Describe  the  angles  A,  180°-  A, 
OP  being  the  terminal  line  of  A,  and  OPi  the  terminal  line 
of  180°-  A.    In  Figs.  70,  71,  72,  73,  A  is  an  angle  in  the 
first,  second,  third,  fourth  quadrants  respectively. 
px  p     p  px 

'  x   f\^^^1x  M       O       Mi     Mi      o        M 


Mx 


M 


Fig.  70.  Fig.  71.  Fig.  72.  Fig.  73. 

Take  OPx  =  OP,  and  draw  PM,  PxMi  at  right  angles  to 
OX.     Then  in  each  figure 

OMx  =  -  OM,  MtPi  =  MP. 

Hence,  for  any  angle  A, 


MiPt 


sin  (180°- 4)  = 
cos  (180°-  A) 


OPi 
0MX 


MP 

OP 

OM 


=     sin  A 


=  —cos  A. 


OPx        OP 

So  also, 

tan  (180°-  A)  =  -tan  A;    cot  (180°-  A)  =  -cot  A; 

sec  (180°-A)  =  -sec  A;     esc  (180°-  A)  =     esc  A. 

Thus,  a  function  of  the  supplement  of  an  angle  and  the 
same  function  of  the  angle  itself  are  numerically  equal;  the 
sines  and  cosecants  respectively  of  supplementary  angles  have 
the  same  sign,  but  the  cosines  and  the  remaining  functions 
respectively  have  opposite  signs. 

Functions  of  180°  +A.  On  proceeding  as  in  the  pre- 
ceding cases  it  will  be  found  that 

sin  (180°  +A)  -  -  sin  A,  cos  (180°  +  A)  =  -  cos  A, 
tan(180°  +  A)=  tan  A,  cot(180°  +  A)=  cot  A, 
sec  (180°  +  A)  =  -sec  A,     esc  (180°  +  A)  =  -esc  A. 


66  ELEMENTS  OF  PLANE  TRIGONOMETRY 

Since,  OX  being  the  initial  line,  all  angles  having  the  same 
terminal  line  have  the  same  value  for  a  function,  it  follows 
that 

any  function  of  360°  +  A,  and  of  rc-360°  +  A  (n  =  any 
whole  number)  is  equal  to  the  same  function  of  A. 
Also, 

any  function  of  360°—  A,  and  of  n-360°—A  (n  =  any 
whole  number)  is  equal  to  the  same  function  of  —A. 

EXAMPLES 

1.  Express  the  functions  of  270°— A  in  terms  of  functions 
otA. 

2.  Express  the  functions  of  270°  +  A  in  terms  of  functions 
of  A. 

27.  Reduction  of  trigonometric  functions  of  any  angle 
to  functions  of  acute  angles.  By  means  of  the  relations  in 
Art.  26  the  functions  of  any  angle  can  be  expressed  in  terms 
of  the  functions  of  an  angle  between  0°  and  90°,  and  in  terms 
of  the  functions  of  an  angle  between  0°  and  45°. 

EXAMPLES 

1.  sin  700°  =  sin  (360° +  340°)  =  sin  340°  =  sin  (-20°) 

= -sin  20°= -.3420. 

2.  tan975°  =  tan  (2- 360° +  255°)  =  tan 255°  =  tan  (180°  +  75°) 

=  tan75°  =  cot  15°  =  3.7321. 

3.  esc  (-1160°)= -esc  (1160°)  = -esc  (3-360°  +  S0°) 

= -esc  80°= -sec  10°. 
Hence 

esc  (-1160°)=  -sec  10°=  -_i_--^.=  -1.015. 


ANGLES  AND  THEIR  TRIGONOMETRIC  FUNCTIONS      67 

4.  Express  the  following  as  functions  of  acute  angles: 

(a)    sin  287°.  (d)  sec  925°  10'.        (g)   tan  (-1055°). 

(6)  cos  332°.  (e)   sin  2150°.  (h)  cos  (-2055°). 

(c)  tan  218°  30'.       (/)  cot  (-487°).       (i)    esc  310°  30'. 

5.  Find  the  values  of  the  following: 

(a)  sin  346°  10'.        (d)  sec  (-310°)         (g)  esc  876°. 

(6)  cos231o30/.        (e)   cot  950°.  (h)  cos  (-1131°). 

(c)  tan  174°  15'.        (/)  sin  (-2830°)      (i)  tan  (-1487°). 

6.  Prove  the  following: 

(a)  cos  240°  cos  120°-sin  120°  cos  150°- 1. 
(6)  tan  675°  sec  540°  +  cot  495°  esc  450° =0. 


CHAPTER  IV 

GENERAL   VALUES.     INVERSE   TRIGONOMETRIC 
FUNCTIONS.    TRIGONOMETRIC  EQUATIONS 

28.  General  values.  All  angles  having  the  same  initial 
and  terminal  lines  have  the  same  values  for  each  trigono- 
metric function.  The  general  value  of  an  angle  having  a 
given  trigonometric  function  is  an  expression,  or  formula, 
which  includes  all  the  angles  that  have  that  function. 

General  expression  for  all  angles  having  the  same  sine 
and  cosecant.  Let  a  be  an  acute  angle  having  a  given  sine, 
a,  say.     Let  its  terminal  line  be  OP.    All  angles  having  OP 


o 
Fig.  74. 

for  terminal  line  have  the  same  sine  a.    These  angles  are 
the  angles  in  the  expression, 

ra-360°  +  a,    i.e.,   2m -180°  + a,    (ra  =  any   integer).     {%) 

Also  sin  (180°—  a)  =  sin  a  =  a. 

In  Fig.  74,  OPi  is  the  terminal  line  of  180°- a.  All 
angles  having  OPi  for  terminal  line  have  the  same  sine  a. 
They  are  the  angles  n  the  expression, 

m-360o  +  (180o-a),  i.e.,  (2m  +  l)180°-a,  (m  an  integer),  (it) 

68 


GENERAL  VALUES  69 

Expressions  (i)  and  (it)  are  both  included  in  the  expres- 
sion, 

nl80°+(—  l)na,        i.e.,    mr-f(— l)na,      n  an  integer. 

For,  if  n  is  even,  a  has  the  sign  +,  as  in  (i);  if  n  is  odd, 
a  has  the  sign  — ,  as  in  (u). 

Since  esc  a  =  i ,  this  expression  includes  all  angles  that 

have  the  same  cosecant  as  a. 
Otherwise  expressed: 

sin  a -sin  [rc-180°  +  (—  l)na]  =  sin  [n7r  +  (— l)na]; 

esc  a=csc  [tt-180°  +  (—  l)na]  =  csc  [ft7r  +  (—  l)na]. 

'Hie  proof  is  similar  for  a  in  any  quadrant. 

Vs 

Ex.  1.  Give  the  general  expression  for  a  if  sin  a  =  -^~. 

The  least  positive  value  of  a  is  60°. 

.*.  The   general   value    of   a    is    n-180°+(-l)n  60°,    i.e., 

™r+(-l)|. 

On  giving  n  the  values  0,  1,  2,  3,  ...  ,  particular  values  of 
a  are  obtained,  viz.,  60°,  120°,  420°,  480°,  .  .  . 

General  expression  for  all  angles  having  the  same  cosine 
and  secant.  Let  a  be  an  acute  angle  having  a  given  cosine, 
a,  say.  Let  its  terminal  line  be  OP.  All  angles  having  OP 
for  terminal  line  have  the  same  cosine  a.  These  angles 
are  the  angles  in  the  expression, 

n  -  360°  +  a ,         {n  =  an  integer. )  (m) 

Also  cos  (—  a)  =  cos  a  =  a. 


70 


ELEMENTS  GF  PLANE  TRIGONOMETRY 


In  Fig.  75,  OP i  is  the  terminal  line  of  —a.  All  angles 
having  OPi  for  terminal  line  have  the  same  cosine  a.  They 
are  the  angles  in  the  expression, 

n  •  360°  —  a j         (n  =  an  integer. )  (iv) 

Expressions  (Hi)  and  (iv)  are  both  included  in  the  expres- 
sion, 

n  •  360°  ±  a,         i.e.,  2wir  ±  a,         (n  an  integer.) 

Since   sec  a  = ,    this   expression   includes   all   angles 

that  have  the  same  secant  as  a. 
Otherwise  expressed: 

cos  a  =  cos  (n  •  360°  ±  a)  =  cos  (2m:  ±  a) ; 

sec  a  =  sec  (n  •  360°  ±a)=  sec  (2tt7r  ±  a) . 

The  proof  is  similar  for  a  in  any  quadrant. 


Fig.  75. 


Ex.  2.    Give  the  general  expression  for  a  when  cos  a=J. 
The  least  positive  value  of  a  is  60°. 

.'.  The  general  value  of  a  is  n- 360° ±60°,  i.e.,  2nn±n-. 

o 

On  giving  n  the  values  0,  1,  2,  3,  ...  ,  particular  values  of 

a  are  obtained,  viz.,  ±60°,  420°,  300°,  780°,  660°,  .  .  , 


GENERAL  VALUES 


71 


General  expression  for  all  angles  having  the  same  tangent 
and  cotangent.  Let  a  (Fig.  76)  be  an  acute  angle  having  a 
given  tangent,  a,  say.  Let  its  terminal  line  be  OP.  All  angles 
having  OP  for  terminal  line  have  the  same  tangent  a. 
These  angles  are  the  angles  in  the  expression, 

ra-360o  +  a,     i.e.,  2m -180°  + a,     (m     an     integer.)       (v) 

Also  tan  (180°  +  a)  =  tan  a. 

In  Fig.  76,  OP i  is  the  terminal  line  of  180°+ a.  All 
angles  having  OPi  for  terminal  line  have  the  same  tangent  a. 
They  are  the  angles  in  the  expression 

m- 360°  +  (180°  +  a)  i.e.,  (2m  +  l)180°  +  a,  (m  an  integer)  (vi) 


180°+ a 

\a 

M, 

-1 

\CL 

1 

y^o 

M 

-a{ 

Fig.  76. 

Expressions  (v)  and  (vi)  are  both  included  in  the  expres- 
sion, 

w-180°  +  a,         i.e.,    mr  +  a,         n  an  integer. 

Since  cota=- ,   this  expression  includes  all  angles 

tan  a'  r 

that  have  the  same  cotangent  as  a. 

Otherwise  expressed : 

tan  a  =  tan  (n  •  180°  +  a)  =  tan  (un  +  a) ; 

cot  a=cot  (tt-180°  +  a:)=cot  (nn+a). 

The  proof  is  similar  for  a  in  any  quadrant. 


72  ELEMENTS  OF  PLANE  TRIGONOMETRY 

Ex.  3.  Give  the  general  expression  for  a  when  tan  a  =  \/3. 
The  least  positive  value  of  a  is  60°. 

.'.  The  general  value  of  a  is  n-180°  +  60°,  i.e.,  nn  +—. 

o 

On  giving  n  the  values  0,  1,  2,  3,  ...  ,  particular  values  of 
a  are  obtained,  viz.,  60°,  240°,  420°,  600°,  .  .  . 

EXAMPLES 

4.  Given  sin  A = i,  find  the  general  value  of  A .     Also  find 
the  four  least  positive  values  of  A. 

5.  Given  cos  A  =  —  J,  find  the  general  value  of  A  and  the 
three  least  positive  values  of  A. 

6.  Find  the  general  value  of  0  which  satisfies  each  of  the 
following  equations : 

(a)    sin  0  =  0.  (g)    sin#=  — 


(6)    cos  0  =  0.  (h)   cos  6= ]=. 


2  ' 

V2 


(c)   tan  0  =  0.  (i)    tan  0=1. 

1 

Vs' 


(d)   cot0=-^.  (/)    cot0=-l. 


2 

(e)    sec  0=1.  (k)    sec  0=— -=^. 

V3 

(/)  csc0=\/2.  (I)    tan0= — ~ 

V3 

29.  Inverse  trigonometric  functions.  It  has  been  seen 
that  the  sine  is  a  function  of  the  angle.  On  the  other  hand, 
the  value  of  the  angle  depends  on  the  value  of  the  sine, 
and  thus  the  angle  is  a  function  of  the  sine.    This  function  of 


GENERAL  VALUES  73 

the  sine  is  called  the  inverse  sine  or  the  anti-sine.  Similarly, 
the  angle  is  a  function  of  each  of  the  other  trigonometric 
functions.  Thus  there  arise  other  inverse  trigonometric 
functions  or  anti-trigonometric  functions,  as  they  are  some- 
times called,  viz.,  the  inverse  cosine  or  the  anti-cosine,  the 
inverse  tangent  or  the  anti-tangent,  etc. 

E.g.,  since  sin  30°  =  J, 

30°  is  an  angle  whose  sine  is  J. 

This  line  is  also  expressed  thus : 

30°  is  an  inverse  sine  of  \  (or  an  anti-sine  of  J), 

and  more  briefly,  on  using  a  symbol  invented  for  the  inverse 
sine, 

30°  =  sin"1  J. 

In  general :  the  two  statements, 

"  the  sine  of  an  angle  6  "  is  m, 

6  is  "  an  angle  whose  sine  is  ra," 

are  briefly  expressed, 

0 

6  =  sin-1  m  =  arc  sin  m  * 

The  expressions, 

sin-1  m,  cos-1  m,  tan-1  m,  etc.,  .  .  .  , 

are  the  symbols  for  the  inverse  trigonometric  functions. 
The  expression  tan-1 2,  for  instance,  is  read  "  the  inverse 

*  This  notation  is  the  one  used  on  the  Continent. 


74  ELEMENTS  OF   PLANE  TRIGONOMETRY 

tangent  of  2,"  "the  anti-tangent  of  2,"  and  may  also  be 
read  "  an  angle  (or  the  set  of  angles)  whose  tangent  is  2." 

N.B.  The  trigonometric  functions  are  ratios,  and  thus  are 
pure  numbers;  the  inverse  trigonometric  functions  are  angles. 

For  instance  (see  Art.  28,  Exs.  1,  2,  3), 

\/3 
sin"1  — -=n- 180°+  (-  l)n  60°,  n  an  integer; 

cos-1  -  =  tt-360°±60°,  n  an  integer; 

tan"1  Vs=n- 180° +  60°,  n  an  integer. 

Thus,  while  each  of  the  trigonometric  functions  has  a 
single  definite  value,  each  inverse  trigonometric  function  has 
an  infinite  number  of  values.  The  angle  having  the  smallest 
numerical  value  in  an  inverse  function  is  called  the  principal 
value  of  the  inverse  function. 

E.g.  The  principal  value  of  tan-1  Vs  is  60°; 
The  principal  value  of  sin_1( y=  J  is  —45°. 

When  the  smallest  numerical  value  has  both  positive  and 
negative  signs  the  positive  sign  is  taken;  thus 
The  principal  value  of  cos-1  \  is  60°. 

N.B.  It  should  be  noted  that  the  "  —  1  "  in  the  symbol 
for  an  inverse  trigonometric  function  is  not  an  algebraic 
exponent.    See  Art.  5,  N.B.  in  Examples. 

Thus:  sin-1  m  does  not  denote  (sin  m)~l,  i.e.,  -. ; 

v  '       '  sin  m 

(tan  x)~l,    i.e.,    t ,   should  not  be  written  tan-1  x. 

tan  x 


GENERAL  VALUES  75 


EXAMPLES 
1.  Prove  that  sin-1  £=cos-1  Vl—x2. 
Let  6  =  sin-1  x. 

Then     sin#=£. 


Hence  cos  6  =  Vl  —  sin2  d= Vl  —  x2. 

Thus  0=cos_1  Vl  —  x2;     i.e.,     sin-1  x  =  cos_1  Vl  —  x2. 

2.  Construct  the  following: 

(a)  sin"1!.  (c)   tan"1  (-2).         (e)  sec"1 2. 

(6)  cos-i(-|).        (iQ    cot-i(I).  (/)  csc-i(-0. 

Read  the  following  identities,  and  prove  them : 
3.  sin(tan-  £)=  ±±  7.  «rt|-o«n(±^. 

4.ta„(sin-1g)=±|.         8.  8m-»A=tan-«(±|). 

6.  cot"1 1  =  nn*  +  -.  9.  sin  (  cos"1  -  J  =  tan  I  sin"1  — =  J . 

6.  cos"1!  —  K)=2n7t*±—.    10.  tan-1m=cos_1— r=r-. 

V    2/  3  Vl  +  m2 


30.  Trigonometric    equations.      Trigonometric    identities. 

Trigonometric  equations  are  equations  in  which  one  or  more 
trigonometric  functions  or  inverse  trigonometric  functions 
are  involved.  These  equations  are  true  only  under  certain 
conditions,  viz.,  for  certain  values  of  the  angles.  To  solve 
these  equations  is  to  find  these  values. 

*  n  a  whole  number. 


76  ELEMENTS   OF  PLANE  TRIGONOMETRY 

E.g.  The  equation 

tan  x=l. 
Here  x  =  tan"1  1  =  45,°  225,°  etc. 

The  general,  or  complete,  solution  of  this  equation  is 

z=n-180o  +  45o,     i.e.,     vm+j. 

Trigonometric  identities  are  in  the  form  of  equations  and  are 
unconditionally  true,  i.e.,  are  true  for  all  values  of  the  angles 
involved. 

E.g.  sin2  x  +  cos2  x  =  1, 

and  the  other  relations  in  Art.  25,  (1),  (2),  (3),  are  identities. 

EXAMPLES 

1.  Solve  the  equation 

sin2  x  —  2  cos  x-\- }  =  0. 

Here  1  —  cos2  x  —  2  cos  £-f-J  =  0. 

.'.  4  cos2  x+S  cos  x— 5  =  0; 

i.e.,  (2  cosx  +  5)   (2  cosz-l)=0. 

.*.  2  cos  £  +  5  =  0,  or       2  cos  x— 1  =  0. 

.'.  cosz=  —  f,       or  cos  x  =J. 

There  is  no  solution  for  cosx=— f,  since  the  cosine  of  an 

angle  lies  between  —  1   and    +1.*     From  cos  x  =  J  comes  the 

solution, 

z=tt-360o±60°. 

In  solving  an  equation  containing  several  functions,  the 
general  method  is  to  reduce  the  equation  to  a  form  in  which 
only  one  function  appears. 

*  In  other  words:  The  solution  is  impossible. 


GENERAL  VALUES  77 

2.  Prove  the  identity 

cos4  A  —  sin4  .4  =  1  —  2  sin2  A. 
cos4  A  —  sin4  A  =  (cos2  A  +  sin2  A)  (cos2  ^4  —  sin2  A) 

=  1  •  (1-sin2  A  -sin2  4)  =  1-2  sin2  A. 
Solve  each  of  the  following  equations: 

3.  2  cos2  z  +  5  sinx— 4  =  0.      8.  sin  z=tan2  x. 

4.  sin  z  +  csc  x  =  2.  9.  2  sin2  #  +  V3  cos  rr+l  =  0. 

5.  sin  y  +  cos  y  =  V2.  10.  4  sec2  5-7  tan2  5=3. 

6.  sin2a;=l.  11.  4  tan  x— cot  z  =  3. 

7.  2  cos  A  +  sec  A  =  S.  12.  sec2  ?/— 5  tan  y+5  =  0. 

13.  Given   4  sin2  0  =  3,    find    the    values    of    0   which   are 
between  0°  and  500°. 

14.  Given  sin  z+cos  x  cot  x =  2,  find  the  values  of  x  which 
are  between  0°  and  360°. 

Prove  the  following  identities: 

15.  sin3  0 + cos3  0  =  (sin  0  +  cos  0)  (1  —  sin  0  cos  0) . 

16.  sin  z  (cot  rr+2)  (2  cot  x+l)=2  cosec  x+5  cos  a:. 

17.  cos6  A  +  sin6  A  =  1  —  3  cos2  A  sin2  ^4 . 

18.  cos6  x  +  2  cos4  x  sin2  z-f-cos2  #  sin4  #  +  sin2  #=1. 
sec0  +  csc0     l  +  cot0     tan  0+1 


19. 


sec  0  —  esc  0     1  —  cot  0     tan  0—1' 
sec2  0  +  esc2  0 


20.  tan0  +  cot  0  = 


sec  0  esc  0 


CHAPTER  V 

TRIGONOMETRIC   FUNCTIONS   OF    THE   SUM   AND 
DIFFERENCE  OF  TWO  ANGLES 


General    Formulas 

31.  To  deduce  sin  (A  +  B),  cos  (A  +  B).  N.B.  In  Arts.  31, 
32  the  conventions  (Arts.  16,  19,  20)  regarding  the  signs  of 
angles  and  lines  are  followed.  The  formulas  can  be  derived, 
however,  for  Figs.  77-79  by  using  the  definitions  in  Art.  3. 

Case  L    A  and  B  both  acute. 

In  Figs.  77,  78, 


O      M  N 
Fig.  77. 


M       O     N 
Fig.  78. 


XOL  =  A,    LOT  =  B,    XOT  =  A+B. 


(In  Fig.  77,  A  +B  is  acute;  in  Fig.  78,  A  +B  is  obtuse.) 
From  any  point  P  in  OT,  PM,  PQ  are  drawn  at  right 

angles  to  OX,  OL  respectively;    QN  is  drawn  at  right  angles 

to  OX ;  VQR  is  drawn  parallel  to  OX.    Now 

VQL  =  XOQ  =  A. 


VQP  -  VQL  +  LQP  =  90° + A. 


78 


THE  SUM   AND  DIFFERENCE   OF  TWO  ANGLES  79 

•    /ixln      •    ynP    MP    NQ+KP    NQRP 

NQ 
Now,  QQ=sinA;  ,\  NQ  =  OQ  sin  A. 

7?P 
Also  OP  =sin  H>^  =  sin  (90°  +  A)  =  cos  A. 

.*.  RP  =  QP  cos  A. 

•    /  ^  i  z>\     OQ  sin  A    QP  cos  A 
.-.  sm(A+B)= — Qp  —  + — Qp — . 

But    ~  -  cos  QOP  =  cos  B;    ^p  =  sin  QOP  =  sin  B. 

.\  sm  (A+B)  =sm  A  cos  B  +  qos  Asm  B.  (1) 

fjj-n*  ynp     0M     ON+QR*     ON     OR 

cos(A+B)=cosXOP=Qp= — ^p —  =  qp+qP' 

ON 
Now    -fvj  =  cosA;  .-.  ON  =  OQcos  A 

Also,     7^  =  cos  FQP  =  cos  (90°+A)  =  -sin  A. 

.-.  Qfl=-QPsinA. 

..  OQ  cos  A     QP  . 

.'.  cos(A+£)= — qp — -Q-psmA; 

.'.  cos  (A +  B)=  cos  A  cos  B— sin  A  sin  B.  (2) 

Case  II.     A  and  B  any  angles.     First  suppose  B  acute 
and  A  obtuse,  viz.,  90°  +  A',  A'  thus  being  acute.     Since 

A  -90°+ A', 

A'=-(90°-A). 

*  Note  that  OiV,  Qi2,  are  in  opposite  directions. 


80  ELEMENTS  OF  PLANE  TRIGONOMETRY 

/.  sinA'=-sin(90°-.A)*=-cosA-, 
cos  A'  =  cos  (90°-  A)  *  -  sin  A. 
Then  sin  (A+£)  =  sin  (90°+A'+B)=coa  (A'+B)* 
=  cos  A'  cos  B—  sin  A'  sin  B 
=  sin  A  cos  5 +cos  A  sin  B. 
Also,  cos  (A +^)=  cos  (90°  +  A'+£)  =  -sin  (A'+£)* 
=  —sin  Af  cos  B—cos  A'  sin  5. 
=     cos  A  cos  #— sin  A  sin  B. 

The  above  procedure  shows  that  formulas  (1),  (2)  still 
hold  true  when  one  of  the  angles  A,  B,  in  Case  II,  is  increased 
by  90°,  and  that  these  formulas  will  continue  to  hold  true 
as  the  angles  continue  to  be  increased  by  90°.  Hence  these 
formulas  are  true  for  angles  in  any  quadrants,  and  thus  for  all 
angles. 

Note.  Formulas  (1),  (2)  can  also  be  derived  from  a  figure, 
as  in  Case  I,  for  A  and  B  in  any  quadrants. 

EXAMPLES 


1.  Derive  sin  75°  =  sin  (30°  +  45°)=    ^t} . 


^     .  ^rn     V3-1 

2.  Derive  cos  75°  = =-. 

2V2 

3.  Given  sin#=i,  siny=$,  and  x  and  y  both  acute,  find 
sin  (x+y),  cos  (x+y). 

4.  If  tan  x=  J  and  tan  y  =  -fa,  find  sin  (x+y)  and  cos  (x  +  y) 
when  x  and  y  are  acute  angles. 

5.  Prove  sin  (60°  +  x)  -cos  (30°  +  z)=sin  x. 

*  See  Art.  26. 


THE  SUM  AND  DIFFERENCE   OF  TWO  ANGLES  81 

6.  Prove  cos  (60°+J5)+sin  (30° +B)  =  cos  B. 

7.  Prove  cos  (x  +  y)  cos  z  +  sin  (x  +  y)  sin  #=cos  y. 

8.  Find  sin  (sin-1  J-fsin-1  J)  when  the  angles  are  between 
0°  and  90°. 

9.  Prove  sin  (sin-1  ra+sin-1  n)  =  mv/l- n2±nVl  —  m2. 


10.  Prove  cos  (sin-1  ra  +  sin-1  n)=Vl  —  m2Vl  —  n2±  mn. 

32.  To  deduce  sin  (A—B),  cos  (A—B).  First  method. 
Formulas  (1),  (2),  Art.  31,  are  true  for  all  angles.  On 
taking  —  B  instead  of  B,  there  results : 

sin  (A  —  B)=sinA  cos  (— B)+cos  A  sin  (— B); 

.'.  sin  (A  —B)  =  sin  A  cos  B  —cos  A  sin  B.  (1) 

Also,   cos  (A  —  B)=  cos  A  cos  (— B)~ sin  A  sin  (-B). 

.'.  cos  (^1  — U)  =cos  ^L  cos  B  +  sin  A  sin  ^.  (2) 

Second  method.     Case  I.     A  and  B,  both  acute,  J.  >B. 
In  Fig.  79, 

From  any  point   P  in  077, 
PM,  PQ  are  drawn  at  right  angles  fe 
to  OX,  OL,  respectively;   QN  is 
at  right  angles  to  OX;  VPR  is 
drawn  parallel  to  OX.    Now 


XOL  =  A,    LOT=-B, 
XOT  =  A-B. 


0      N     M 
Fig.  79. 


RQP  =  90°-OQN  =  A; 
.  QPR  =  90°-RQP  =  90°-A; 


82  ELEMENTS  OF   PLANE  TRIGONOMETRY 

.-.  VPQ  =  1$0°-QPR  =  90°  +  A. 
sin  (A  -  B)  =  sin  XOP  =  ~gp=  —gp —  =  Qt>  -  ~jp. 
But    jvpsin  A;  .*.  J\TQ=  30  sin  A 

Also   ~  -  sin  VPQ  -  sin  (90°  +  A)  -  cos  A. 

.'.  RQ  =  PQ  cos  A. 

,a      m     OQsinA     PQcosA 
/.  sin  (A -5)- — ^p— ^p — . 

But        pyp  =  cos  QOP  =  cos  (  —  #)=cos  5; 

PQ    QP 
— YYp  =  ^p  =  sin  QOP  =  sin  (—B)  =  —sinB; 

.'.  sin  (A  —  B)  =  sin  A  cos  B—cos  A  sin  5. 

sj      m  ynP     0M     ON+RP     ON     PR* 

cos  (A - B)=  cos  X0P=Qp  =—jjp — =soP~'5F' 

ON 
Now     7)7)  =  cos  ^;*  •'•  ON  =  0Q  cos  A. 

P7? 
Also      pQ=cos  7PQ  =  cos  (90°  +  A)  =  -sin  A. 

.'.  PR=-PQ  sin  A. 

/,      DX     OQ  cos  A     PQsinA 
.-.  cos(A-£)= — ^p — + — ^p — . 

.*.  cos  (A  —  5)  =  cos  A  cos  B+ sin  A  sin  B. 

*  Since  ltP=-PR. 


THE  SUM  AND  DIFFERENCE  OF   TWO  ANGLES  83 

Case  II.    A  and  B  any  angles.    The  methods  of  proof  are 
similar  to  those  shown  in  Art.  31,  Case  II,  and  Note. 


EXAMPLES 

V3-1 


1.  Derive  sin  15°= sin  (45°  -30°)  = 

2.  Derive  cos  15°= cos  (60° -45°)  = 


2\/2  * 

V3+1 

2V2  ' 


3.  If  cos  A  =  |f  and  cos  B  =  %,  find  sin  (A  —  B)  and  cos  (A  —  B) 
when  A  and  B  are  acute  angles. 

4.  If  tan  x=j  and  tan  y  =  ^,  find  sin  (x  —  y)  and  cos  (x  —  y) 
\  ]  m  x  and  y  are  acute  angles. 

5.  Find  sin  (sin-1  ^  — sin-1  i)  when  the  angles  are  acute. 

Verify  the  following  identities: 

6.  cos  (^+45°)+  sin  (A- 45°)=  0. 

7.  cos  (30°  +  x)~  cos  (30°  -  x)  =  -  sin  z. 

8.  cos  (x  +  y)  cos  (z  —  y)  —  cos2  z  —  sin2  y. 

9.  sin  (x  +  y)  sin  (x  —  2/)  =  cos2  2/  —  cos2  x. 

„rt    sin  (4  +B)  .  ,  .       D 

10.  1 ^  =  tan  A±tan  J5. 

cos  A  cos  B 

33.  Fundamental  formulas.  Formulas  (1),  (2),  Art.  31, 
are  called  the  addition  formulas  or  theorems  in  trigonometry ; 
formulas  (1),  (2),  Art.  32,  are  called  the  subtraction  formulas 
or  theorems.  These  four  formulas  are  also  called  the  funda- 
mental formulas  of  trigonometry.  For  convenience  they  are 
brought  together: 


84  ELEMENTS  OF  PLANE  TRIGONOMETRY 

sin  (A  +  B)  =  sin  A  cos  B+ cos  A  sin  B.  (1) 

sin  (A  —B)  =sin  A  cos  B  —cos  ^  sin  B.  (2) 

cos  (-4  +  B)  =  cos  ^t  cos  B  —sin  ^4  sin  B.  (3) 

cos  (4—  B)=cos  A  cos  JB+sin  ^4  sin  B.  (4) 

7n  words :  Of  any  two  angles, 

sine  sum  =  sine  first  •  cosine  second     +  cosine  first  -  sine  second 
sine  difference  =sine  first  -  cosine  second     —  cosine  first  -  sine  second 
cosine  sum  = cosine  first-  cosine  second— sine  first-  sine  second 
cosine  difference = cosine  first-  cosine  second+sine  first -sine  second 

34.  To  deduce  tan  (A  +  B),  tan(A-B),  cot  (A  +  B),  cot 

(A-B). 

sin  (A  +  B)     sin  A  cos  B  +  cos  A  sin  B 


tan(A+£)  = 


cos  (A +2?)     cos  A  cos  B— sin  A  sin  i? 


On  dividing  each  term  of  the  numerator  and  the  denom- 
inator of  the  second  member  by  cos  A  cos  B  there  is  obtained 

tan  A  +  tan  B 
tan  (.*+*)  =r_tan^tanj?.  (1) 

In  a  similar  way  it  can  be  shown  that 

tan  A  —tan  B  s  v 

tan  (A  -B)  =  -— — -.  (2) 

1  +  tan  A  tan  B  v  J 

•t  .     ^     cos  (A  +B)     cos  A  cos  B— sin  A  sin  B 

COt   (A  +  J5)  = ,   a     t     r>\  =— A d"! J—' D- 

sin  (A  +B)     sin  A  cos  5 +cos  A  sin  B 

On  dividing  each  term  of  the  numerator  and  the  denom- 
inator of  the  second  member  by  sin  A  sin  B  there  is  obtained 


cot  A  cot  B— 
cot  B  +  cot 

In  a  similar  way  it  can  be  shown  that 


cot(A+i?)=cotjS+cotA.  (3) 


THE   SUM  AND  DIFFERENCE  OF  TWO  ANGLES  85 

j.  ,  A        r»N       COt  ^4  COt  5  +  1 

cot^-^°cotg-cot^-  (« 


EXAMPLES 

1.  If     tanP=2,     tanQ  =  $,     show     that     tan(P+Q)  =  7, 
tan(P-Q)  =  l. 

2.  Derive  tan  75°  from  tan  45°  and  tan  30°. 

3.  Derive  tan  15°  from  tan  60°  and  tan  45°. 

4.  Prove:  tan  (45°  +  x)  =  \  +  ^n  X ;  tan  (45°-x)  =  t1~tan  x. 

'     1-tanz'  7     1  +  tanz 

5.  Prove:  cot  (45°  +  x)  =  ^t^li ;  cot  (45°-:r)  =  Cot  *+1 

cotx+T  COtx— 1 

6.  If  tan  A  =  £,  tan  5=  J,  find  tan  (A +  5)  ani  tan  (A  —  B). 

7.  Show  that  tan"1  m+tan"1  n  =  ta.nrl  m+^. 

1  —  mn 

Let  x  =  tan_1  ra,     and     2/=tan-1n. 

Then  tanx  =  ra,  tany=n. 

„  /    ,    \       tanx  +  tan?/       m-\-n 

Now     tan  (x  +  y)=  - — — ■*- = - — — . 

1  —  tan  x  tan  y     1  —  mn 

.'.  £  +  v=tan  *- ;     i.e.,     tan  1m+tan  1n=tan~1- . 

9  1  — ran'  1  —  mn 

8.  Show  that     tan-1  m  —  tan-1  n  =  tan-1 


1  +  mw 

9.  Find  tan-1  7  + tan-1  3,  and  tan-1  7  — tan-1  3. 

10.  Find  tan-1  2  + tan-1 -5,  and  cot-1  2  + cot-1 -5. 

35.  To  deduce  sin  2-4,  cos  2A,  tan2A.     On  putting  B  =  A 
in  formulas  (1),  (3),  Art.  33,  there  results: 


86  ELEMENTS  OF  PLANE  TRIGONOMETRY 

sin  2A  =  2  sin  A  cos  A;  (1) 

cos  2A  =  cos2  A  -  sin2  A ;  (2) 

i.e.,  =1-2  sin2  A*;  (3) 

i.e.,  =  2  cos2 .4-1*  (4) 

On  putting  B  =  A  in  formula  (1),  Art.  34: 

2  tan  A 
tan2A  =  l^U^A-  ® 

On  putting    2^4.  =  z;     then     A  =  $x;  and   these  formulas 
are  expressed : 

sin  x  =  2  sin  Jz-cos  \x, 

cos  z  =  cos2  Jo;— sin2  }x, 

etc. 
In  words : 

sine  any  angle  =  2  sine  half-angle  •  cosine  half-angle, 

cosine  any  an#Ze  =  (cosine  half-angle)2—  (sine  half-angle)2, 

=  1  —  2  (sine  half-angle)2, 

•=  2  (cosine  half-angle)2  —  1 . 

.  2  tangent  half-angle 

tangent  any  «^_T_^____. 

EXAMPLES 
1.  Find  cos  22£°  from  cos  45°. 

2cos222£°=l  +  cos45°. 

•    co3222A°-Vli      M- 1  +  V?     1  +  1-4142 
..  cos  22*  -2^  +  V2J— 2VF"2X1.4142  =  -8536' 

.\  cos  22i°=.9239. 

♦Sinoecos2   ^  +  sin2A  =  l. 


THE   SUM  AND  DIFFERENCE  OF  TWO  ANGLES  87 

2.   (a)  Deduce  sin  22^°  from  cos  45°. 

(b)  Deduce  tan  22^°  from  tan  45°, 

(c)  From  functions  of  180°  derive  sin  90°,  cos  90°,  tan  90°. 
cot2A-l 


3.  Derive  cot  2A 


2  cot  A 


3 

4.  Express  cos  3x  and  sin  3x  in  terms  of  functions  of  —x. 

5.  Express  cos  3x  and  sin  3x  in  terms  of  functions  of  6x. 

6.  Express  cos  6x  and  sin  Qx  in  terms  of  functions  of  3x. 

7.  Derive  the  following: 


.     ,          -  /l  —cos  oc                               /l  +COS  00 
sin  §«  -  \ — — ;       cos  fcc  =  \ ; 


,           /l  -cos  oc 
tan  Jos  —  \— . 

11  1  +  COS  oc 

8.  Derive  the  following: 

sin  3A  =  3  sin  A  —4  sin3  .4; 

cos  3A  =4  cos3  ^4  —3  cos  A\ 

'    A     3  tan  A  —  tan3  ^4 
tan  3^4  =  — - — — — x— - — . 
1—  3  tan2  A 

[Suggestion :  sin  3 A  =  sin  (2 A  +  A) 

=  sin  2A  cos  A  +  cos  2 A  sin  A  =  etc] 

9.  Verify  the  following  identities: 

(1)  cot  A-cot2A=cosec2A.    (2)  1  +tan2A  tan  A  =  sec  2A. 

^  /  .   A  ,        M2     1  ,    •     ,    /.x      sin2A        .        , 
V3)  (sm-±cos-j  =l±smi.  (4)  1  +  CQs2^tan  A 


88  ELEMENTS  OF   PLANE  TRIGONOMETRY 

/KX  1  —  cos  24  .  /fi,.l  +  cos4         x4 

(5)     gl-    oj     =  tan4.  (6)  - — -. — -r— =  cot-. 

sin  24.  N        smi  2 

_     2  tan  4         .  /ox2— sec24  _  . 

(7)  rTT — 2-r  =  sm  2^-  (8) 9-r-=cos  24. 

1  +  tan2  4  sec2  4 

(9)  cos4  0-sin  40= cos  20.         (10)  cot  0-tan  0  =2  cot  20. 

/1ixsin20     cos  20  a  ,•    .  sin3x     cos  3a;     _ 

(11)  — r—z ^-  =  sec  0.  (12)  — s =  2. 

sin  0       cos  0  sin  #      cos  x 

1-tan2^ 

/io\  /•  ^  fiA\  j.      x        smi        1  —  cos  a: 

(13)  cos  4  = r.  (14)  tan— =  - = — : . 

1     ,     2  4  2     1  + cos  a;       sini 

^.v  J       .         4  tan  a:  —  4  tan3  x 

(15)  tan4x=- — — — = — — — j— . 

1  —  6  tan2  x  +  tan4  a; 

(16)  sin  4a:  =  4sina:cosa:  —  8  sin3  a:  cos  a; 

=  8  cos3  x  sin  a;  — 4  cos  x  sin  x. 

(17)  cos  4x=l  —  8  sin2  z  + 8  sin4  2=1  —  8  cos2  a: +8  cos4  a;. 

4 
10.  (a)  If  sin  4=-,  calculate  cos  4,  sin  24,  cos  24,  tan  24. 

2  1 

(6)   If  cos  24=—,  prove  tan  4=—  \/~b. 
o  o 

1  4 

(c)   If  tan  4  =-,  prove  cos  24  =  +  -. . 
o  5 

36.  Transformation  formulas.  From  formulas  (l)-(4), 
Art.  33,  there  follow,  on  making  the  add  tions  and  sub- 
tractions indicated : 

sin  (^  +  -B)+sin  (A—B)  =     2  sin  A  cos  B.  (1) 

sin  (A  +  B)  -sin  (A  -B)  =     2  cos  A  sin  B.  (2) 

cos  (4  +  J3)  +  cos  (A—B)=     2  cos  A  cos  B.  (3) 

cos  (A  +  B)  -cos  (A  -B)  =  -2  sin  A  sin  ^.  (4) 


THE  SUM  AND  DIFFERENCE  OF  TWO  ANGLES  89 

By  this  set  of  formulas  products  of  sines  and  cosines  can 
be  transformed  into  sums  and  differences.    Thus: 

In  words  (reading  the  second  members  first):  Of  any 
two  angles, 

2  sin  one  •  cos  the  other  =  sin  sum  +  sin  difference,*    (1/) 

2  cos  one  •  sin  the  other  =  sin  sum — sin  difference,      (2') 

2  cos  one  •  cos  the  other  =  cos  difference  +  cos  sum,       (3') 

2  sin  one  •  sin  the  other  =  cos  difference — cos  sum.      (4') 

On  putting  A+B  =  P, 

A~B  =  Q, 

and  solving  for  A  and  B, 

A  =  UP  +  Q),      B  =  i(P-Q). 

■  Formulas  (l)-(4)  then  take  the  forms: 

sin  P  +  sin  Q=     2  sin  — r~  cos  .  (5) 

sin  -P— sin  Q=     2  cos     9     sin  .  (6) 

COS  -P  +  COS  Q  -       2  COS  ^2  cos  *!z2  (7) 

cos  1* -cos  £  =  -2  sin  ^-^  sin  ^2  (8) 

By  this  set  of  formulas  sums  and  differences  of  sines  and 
cosines  can  be  transformed  into  products. 
In  words:    Of  any  two  angles, 

the  sum  of  the  sines =2  sin  half  sum -cos  half  difference,*      (5') 
the  difference  of  the  sines =2  cos  half  sum- sin  half  difference,         (6') 
*  The  difference  is  taken:  first  angle -the  second. 


90  ELEMENTS  OF  PLANE  TRIGONOMETRY 

the  sum  of  the  cosines =2  cos  half  sum  •  cos  half  difference,         (70 
the  difference  of  the  cosines  =  —  2  sin  half  sum  ■  sin  half  difference.     (8') 

EXAMPLES 

a     ai_         xi.    x  COS  X—  COS?/ 

1.  Show  that ■ ^=  —tan  J  (z+y)  tan  J(s— y). 

COS  Z  +  COS  i/ 

cos  a:  — cos  ?/_  —2  sin  \(x-\-y)  sin  J(^~2/) 
cos  z  +  cos  y      2  cos  J(x  +  ?/)  cos  %(x  —  y) 
=  —tan  J(^  +  2/)  tan  %(x— y). 

2.  Show  that  2  sin  (A +45°)  sin  {A  -45°)=  sin2  A  -cos2  A. 
2  sin  (A +  45°)  sin  {A  -45°) 


=. cos  ( A  +  45°  -  A  -  45°)  -  cos  (A  +  45°  +  A  -  45°) , 
=  cos  90° — cos  2 A  =  sin2  A  —  cos2  A . 

«    ot_       xu   .sin  A  +  sin  3A. 

3.  Show  that r- — -t  =  tan  2A. 

cos  A  +  cos  3A 

sin  A  +  sin  3A  Jl  sin^(3A  +  A)  cosK3A-A) 
cos  A  +  cos  3A  ~2  cos  i(3A  +  A)  cos  i(3A  -  A)1 

sin  2A  • 

= pr-r=tan  2A. 

cos2A 

4.  Solve  the  equation     sin  50  +  sin  0  =  sin  30. 

.'.  2  sin  30  cos  20  =  sin  30.       .'.  sin  30(2  cos  20-1)=  0. 
/.  (a)  sin  30=0;       (6)  2  cos  20-1  =  0. 

From  (a),  30=0°,  180°,  etc.;   the  general  value  of  30  is  nn 
(n  being  any  integer). 

.*.  0  =  0°,  60°,  etc.;  the  general  value  of  0  is  — . 

o 

From  (b),  cos  20  =  £. 

.\  20=  ±60°,  etc.;  its  general  value  is  2rwr±— . 

o 

.*.     0=  ±30°,  etc.;  its  general  value  is    nn±— . 

o 


THE  SUM  AND  DIFFERENCE  OF   TWO  ANGLES  91 

5.  Transform  each  of  the  following  sums  and  differences 
into  a  product: 

(1)  sin  3x  +  sin  bx.  (2)  sin  7A  —  sin  5A. 

(3)  cos2z— cos  6z.  (4)  cos  bx  +  cos  9x. 

(5)  sin  mA  +  sin  nB.  (6)  cos  mx— cos  ny. 

(7)  sin  Sx  +  cos  bx.  (8)  cos  4x— sin  2x. 

6.  Transform  each  of  the  following  products  into  a  sum  or 
a  difference : 

(1)  sin  5x  cos  Sx.  (2)  cos  7x  sin  bx. 

(3)  sin  2z  sin  Qx.  (4)  cos  bx  cos  9x. 

(5)  sin  mA  sin  nB.  (6)  cos  nx  cos  my. 

(7)  sin  4z  sin  2x.  (8)  cos  7z  cos  Sx. 

7.  Show  that  the  value  of 

sin  (n  + 1)5  sin  (n-l)£+cos  (w+ 1)5  cos  (n-l)B 

is  independent  of  n. 

Verify  the  following  identities: 

8.  (a)  sin  (w+l)A  +  sin  (n—  1)A  =  2  sin  nA  cos  A. 
(6)   cos  (n+l)A  +  cos  (n—  1)A=2  cos  nA  cos  A. 

9.  cos  (A  +  5)  cos  A  +  sin  (A  +  5)  sin  A  =  cos  5. 

10.  esc  2 A  +  cot  2 A  =  cot  A . 

11.  sin  5 A  sin  A  =  sin2  3 A  —  sin2  2A. 

..    sin  Sx— sin  x  „„    sinA  +  sini?     tan  £(A  +  Z?) 

12. 5— =  tanz.    13.  -^ — rX-, — 5=- \)  .        ' 

cos  3x+cos  x  sin  A  — sin  5    tan  i(A  —  5) 

sin  (a?+y)_ tan x  +  tan  ?/  cos  (x  +  y) __\  —  tan x tan ?/ 

sin  (x—  y)     tana;— tan?/'  '  cos  (x— y)     l  +  tanxtany' 

16.  sin  Sx + sin  bx  =  8  sin  x  cos2  x  cos  2x. 

17.  cos  20°  cos  40°  cos  80°  =  .125  (without  use  of  tables). 


720 
1681 


92  ELEMENTS  OF  PLANE  TRIGONOMETRY 

18.  Given  tan  A=-,  t&nB=-,  tanC=^,  find  tan  (A+B+C), 
Prove  the  following: 

19.  sin  \2  tan-!^j  = 

20.  sin[isin-i(-|)]  =  ±|    and     ±i 

21.  3  sin-1  a:=sin_1  (3x— 4x3). 

22.  3  cos-1  x=cos_1  (4:X3-3x).  23.  sec-1  3  =  2  cot-1  V2. 

24.  tan"1*  +  tan"1  y  +  tan"1* = tan"1 5±y±£zfSL. 

l  —  xy  —  yz—zx 

25.  sm-i-  +  sm-i-  +  sini-  =  -. 

-1   a  —  b   ,  x     -1   b  —  c  -   c—a 

26.  tan  1— — r  +  tan  1  1  ,  ,   +tan  J  — — =0. 

1  +  ao  1  +  6c  1  -f  ca 

n„    T-A      A     a     ,        ^u       .     .        2a6      .    _  .      ±4a6(a2-62) 

27.  If  tan  £-  =  r,  snow  that  sin  A  —  0  ,  ,  9.  sm  2 A  =  — ,  „  ,  LOx0     . 

2  6  a2  +  o27  (a2  +  62)2 

Solve  the  following  equations: 

28.  cos  0 -cos  70  =  sin  40.  29.  sin  20  +  sin  40  =  V2  cos  0. 
30.  sin  2a  +  cos  2a  =1.                   31.  sin  2a +  2  cos  2a  =  1. 

32.  sin  20  +  2  sin  40  +  sin  6(9  =  0.    33.  4  sin  0  cos  20=1. 

34.  tan-1  2^  +  tan-13x=— .      35.  cos-1£— aia-1  x=cos~1  x\/3. 

o 

36.  tan"1  (z+l)+tan-1  (x—  l)=tan_1  — . 

ol 

37.  Find  tan  (A  +  B+C)  in  terms  of  tan  A,  tan  B,  tan  C. 
Thence  show: 

(a)  If  A+£+C=  90°, 


THE  SUM  AND  DIFFERENCE  OF  TWO  ANGLES  93 

tan  A  tan  2?  +  tan  B  tan  C+tan  Q  tan  A  =  l; 
(6)  If  A  +  B  +  C=  180°, 

tan  A  +  tan  B+ tan  C=tan  A  tan  B  tan  C. 
38.  Prove  the  following,  given  that  A  +  B  +  C=  180°: 

A        B       C 

(a)  cos  A  +  cos  B  +  cos  C=l+4  sin—  sin  —  sin—; 

A        B        C 

(b)  sin  A+sin  5+sin  C=4  cos  —  cos  —  cos  — . 


CHAPTER  VI 

RELATIONS    BETWEEN    THE    SIDES    AND    ANGLES    OF    A 

TRIANGLE 


37.  Notation.     Simple  geometrical  relations. 

Notation: 

In  stating  and  deriving  the  relations  in  Arts.  37-41  the 
triangle  is  denoted  as  ABC,  and  the  sides  opposite  the  angles 
A,  B,  C,  as  a,  b,  c,  rsepectively. 

Simple  geometrical  relations : 

(a)  A  +B  +  (7=180°. 

(6)  The  greater  side  is  opposite  the  greater  angle, 
and  conversely. 

38.  The  law  of  sines.  From1  C  in  the  triangle  ABC 
draw  CD  at  right  angles  to  opposite  side  AB,  and  meeting  AB 
or  AB  produced  in  D.     (In  Fig.  80  B  is  acute,  in  Fig.  81 

o  ~ — ^o 


D  BV 

Fig.  80. 


A    c    BD  V 
Fig.  82. 


B  is  obtuse,  and  in  Fig.  82  B  is  a  right  angle.)     Produce  AB 

to  V.     In  what  follows,  AB  is  taken  as  the  positive  direction. 

94 


THE  SIDES   AND  ANGLES  OF  A  TRIANGLE  95 

In  CD A,  DC  =  b  sin  A. 

In  CDB  (Figs.  80,  81), 

DC  =  a  sin  VBC 

=  a  sin  B. 

In  Fig.  82,  DC=BC=a=a  sin  B  * 

Therefore,  in  all  three  triangles, 

a  sin  B  =  b  sin  A. 

Tj  a  °  ,     a     sini 

Hence,  - — -r  =  - — B,     and     r  =  ~ — ™-  (lj 

sin  A     sin  B'  b     sm  B  w 

Similarly,  on  drawing  a  line  from  B  at  right  angles  to 
AC,  it  can  be  shown  that 

a  c  ^     a     sin  A 

- — T=-r— 7?,     and     -— - — jy.  (2) 

sm  A     sm  C  c     sm  C  v 

Hence,  in  any  triangle  ^4.5(7, 

a  b  c 


-^— =  — — .  (3) 

sin  A     sin  B     sin  O*  v  ' 

In  words:  The  sides  of  any  triangle  are  proportional  to 
the  sines  of  the  opposite  angles. 

The  circle  described  about  ABC.  Each  fraction  in  (3) 
gives  the  diameter  of  this  circle.  Let  0  (Fig.  83)  be  its  centre 
and  R  its  radius.  Draw  OD  at  right  angles  to  any  side, 
say  AB.    Then 

AD  =  \c,  AOD  =  C, 

and  AD  =  AO  sin  AOD;    i.e.,     \c  =  R  sin  C. 

Hence  <IB  =  -^—.  (4) 

sin  C  v 

*  V     £  =  90°,  and  sin  90°=  1. 


96 


ELEMENTS  OF   PLANE    TRIGONOMETRY 


39.  The  law  of  cosines.  The  angle  A  is  acute  in  Fig. 
84,  obtuse  in  Fig.  85,  right  in  Fig.  86.  From  C  draw  CD 
at  right  angles  to  AB.     The  direction  AB  is  taken  as  positive. 

In  Figs.  84,  85, 

BC2  =  DC2+DB2. 

In  Fig.  84,    DB  =  AB-AD; 

in  Fig.  85,  DB=DA+AB=-AD+AB. 

Hence,  in  both  figures, 

~BC2=DC2  +  (AB-AD)2 

=  DC2+AD2+AB2-2AB.AD. 
In  Fig.  84,    AD  -  AC  cos  B AC; 
in  Fig.  85,      ^   AD  =  AC  cos  BAG  (Art.  20). 
Also, 


DC2 + AD2  =  AC2. 

c               cr^ 

C 

xh    \  *v^ 

b 

^S^rt 

X   •    !  \           i      \     ^ 

Sfc 

jk 

A               D     B             DA0 

Fig.  84.                          Fig.  85. 

Hence,  in  both  figures, 

B 

Fig.  86. 

(1) 


BC  -AG  +AB  -2AG-ABco8A; 

that  is,  a2  =  b2  +  c2  —  2bc  cos  A. 

This  formula  also  holds  for  Fig.  86;  for  there, 

cos  4 = cos  90° =0. 
Similar  formulas  for  b,  c,  can  be  derived  in  like   manner, 
or  can  be  obtained  from  (1)  by  symmetry,  viz. : 

b2  =  c2+a2-2ca  cos  B,     c2  =  a2-\-b2-2ab  cos  C. 


THE  SIDES  AND  ANGLES  OF  A  TRIANGLE  97 

In  words:  In  any  triangle,  the  square  of  any  side  is  equal 
to  the  sum  of  the  squares  of  the  other  two  sides  minus  twice 
the  product  of  these  two  sides  multiplied  by  the  cosine  of  their 
included  angle. 

Relation  (1)  may  be  expressed  as  follows: 

b2+c2-a2 
C0SA  =  -^^'  » 

Similarly: 

■n    c2+a2-b2  „    a2  +  b2-c2 

COS  B  = = ,  COS  C  = tr-r . 

2ca      '  2ab 

40.  The  law  of  tangents.  In  any  triangle  ABC,  for 
any  two  sides,  say  a,  b, 

a  __  sin  A 
b  ~  sin  B' 

From  this,  on  composition  and  division, 

a—b    sin  A  —  sin  B 
a  +  &~sin^4.+sini? 

2  cos  j(A+B)  sin  j(A-B) 
=  2smi(A+B)cosi(A-By      ^  36'  (5)'  ^ 

a_b    tanjU-B) 
*./   «+ft"tanJU+B)-  [Art.  25,  A,  B.]         (1) 

In  words :  The  difference  of  two  sides  of  a  triangle  is  to 
their  sum  as  the  tangent  of  half  the  difference  of  the  opposite 
angles  is  to  the  tangent  of  half  their  sum. 

Now    A+B  =  180°-C.  .\    }(A+5)=90°~ 

/.    tan  i(A  +B)  =  tan  f  90°- ^  )  =  cot  ^ . 
Hence,  relation  (1)  may  be  expressed: 


98  ELEMENTS  OF    PLANE  TRIGONOMETRY 

taniU-B)  =  ^cotlC. 

41.  Functions  of  the  half -angles  of  a  triangle  in  terms  of 
its  sides.  Let  s  denote  half  the  sum  of  the  sides  of  the  triangla. 
Then 

2s  =  a  +  6+c, 


and                  2s—  2a  =  2(s—  a)  =  —  a  +  6+c. 

Similarly,                       2(s-b)=a-b  + 

c, 

2(s-c)=a+b- 

c. 

By  Vrt.  39  (2),                cos  A  =  — ^ 

-a2 

B>   \rt.  35,                2  sin2  |^  =  1 -cos 

A, 

2  cos2  iA  =  l  +cos 

A. 

.*.     2  sin2  J.4                             ".-.     2cos2JA 

i     62+c2  —  a2 
26c 

,     62+c2-a2 
~1+       26c 

26c-62-c2+a2 

26c+62+c2-a2 

26c 

26c 

a2-(6-c)2 

(6+c)2-a2 

26c 

26c 

(a— 6+c)(a  +  6  —  c) 

(6+c+a)(6+c- 

•a) 

26c 

26c 

2(s-6)-2(s-c) 

2s-2(s-a) 

26c       * 

26c 

*"               be 

(1) 

cos2L4-8(8-a>. 
*           be 

(2) 

Since  tan2  \A  =  sin2  JA-f-cos2  \A,  it  follows  that 


THE  SIDES  AND  ANGLES  OF  A  TRIANGLE  99 


lwt»i*-.»-,ft/«-«>.  (3) 

2  sKs-a)  vu' 


■-    sia^=V^^-'    ""^^Fl  (4) 

Similar  formulas  hold  for  %B,  %C,  viz.: 

sin2  \B  = :  cos2  \B  = ; 

ac  ac     ' 

•  2  in     (s-a)(s-b)  s(s-c) 

sin2  J(7  =  - 1 •  cos2  W  =  -^—r — . 

2  ab  2  ab 

(s—a)(s—c) 


tan2  J£  = 


tan2  iC  = 


s(s-6)      ; 

(s— a)(s— 6) 

s(s  —  c) 


Formula  (5)  can  be  given  a  more  symmetrical  form. 
For,  on  multiplying  the  numerator  and  denominator  in  the 
second  member  of  (3)  by  (s—a), 


(s—a)(s—b)(s—c)  _ 
s(s— a)- 


tan2  \A  — 


whence 


tan  \A  =  -1-  J.-"'-*><—\  (6) 

•*        s  -  a  \  8 


T,                                              Is— a  <s-b):s-c)  n\ 

If  r=-\ s »  ''' 

then  tanjU  =  -^.  (8) 

T 

Similarly,        tan  \B  =  — r,     tan  §C  =  — - . 
SO  &>     c 


100 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


The  circle  .nscribed  in  ABC.  The  r  in  the  formulas  above 
is  equal  to  the  radius  of  this  circle.  In  Fig.  87  0  is  the 
centre  of  the  circle,  and  L,  M,  N, 
its  points  of  contact  with  the  sides. 
Let  r  denote  the  radius. 
By  geometry, 

AN  =  MA,    BL  =  NB,     CM  =  LC. 

.-.  AN+BL+LC  =  s,  i.e.,  AN+a  =  s. 

.'.    AN  =  s—a. 


NO 

Now  tan  }A  =  tan  NAO^-r^^ 


AN    s—a' 


(9) 


Comparison  of  (8)  and  (9)  shows  that  the  r  in  (8)  must 
have  the  same  value  as  the  r  in  (9).     Accordingly, 


radius  r 


-V— 


—  a)(8  —  b)(8  —  c) 


EXAMPLES 

1.  Prove:    (a)  with  a  figure,  (6)  without  a  figure,  that  in 
any  triangle  ABC, 

a=b  cos  C  +c  cos  B ; 

and  write  corresponding  formulas  for  b  and  c. 

[Suggestion:    Draw  the  perpendicular  to  BC  or  BC  pro- 
duced.] 

2.  If  the  sines  of  the  angles  of  a  triangle  are  in  the  ratios 
of  13:14:15,  prove  that  the  cosines  are  in  the  ratios  39:33:25. 


3.  In  ABC,  if  a :  b :  c = 8 : 7 : 5,  find  the  angles. 


( 


4.  The  sides  of  a  triangle  are  proportional  to  the  numbers 
4,  5,  6;  find  the  least  angle. 


5.  Prove  that  a  cos  B  —  b  cos  A  = 


ir~ 


CHAPTER    VII 

SOLUTION  OF  OBLIQUE  TRIANGLES 

42.  Cases  for  solution.  General  remarks  on  methods  of 
solution.  In  order  that  a  triangle  may  be  constructed, 
three  elements,  one  of  which  must  be  a  side,  are  required. 
Hence,  there  are  four  cases  for  construction  and  solution, 
namely,  when  the  given  parts  are  as  follows: 

I.  One  side  and  two  angles. 

II.  Two  sides  and  the  angle  opposite  to  one  of  them. 

III.  Two  sides  and  their  included  angle. 

IV.  Three  sides. 

Careful  attention  should  now  be  paid  to  the  remarks 
in  Art.  8  on  the  methods  of  solution  of  triangles  and  to  the 
general  suggestions  made  there  for  solving  problems  and 
checking  results.  These  remarks  and  suggestions  apply 
also  to  the  problems  in  this  chapter. 

Oblique  triangles  can  be  solved  (by  computation)  in 
the  following  ways : 

(a)  By  dividing  them  conveniently  into  right-angled 
triangles,  solving  these  triangles,  and  combining  the  results. 

This  method  is  not  discussed  here,  but  is  left  as  an 
exercise  for  the  student.  Full  details  are  in  Murray,  Plane 
Trigonometry,  Art.  34. 

(6)  By  means  of  certain  relations  in  Chap.  VI,  logarithms 

not  being  used. 

101 


102  ELEMENTS  OF   PLANE   TRIGONOMETRY 

(c)  By  means  of  certain  relations  in  Chap.  VI,  logarithms 
being  used. 

Cases  I-IV  are  solved  in  manner  (b)  in  Arts.  43-46; 
these  cases  are  solved  in  manner  (c)  in  Arts.  48-50. 

43.  Case  I.  Given  one  side  and  two  angles.  In  triangle 
ABC,  suppose  that  A,  B,  a  are  known;  it  is  required  to 
find  C,  b,  c.     In  this  case  (see  Fig.  80,  Art.  38), 

C  =  1S0°-(A+B); 
a  a 


- — n  =  - — j,  whence     b  =  - — T-smB: 

sin  B    sin  A'  sin  A            ' 

_ — —  =--. — -  whence     c  =  - — j -sin  C. 

sin  C     sin  A'  sin  A 


Checks:    a2  =  b2+c2  —  2bc  cos  A ;    — — ^  =  - — j ;    or  other 

'    sin  B     sin  A ' 

results  in  Chap.  VI  which  have  not  been  used  in  the  solution. 

EXAMPLES 

1.  Solve  the  triangle  PQR,  given: 

PQ=  12  in.,  Solution:  *  R  = 

Q  =  40°,  PR= 

P=75°.  RQ  = 

R=  180°-  (P+Q)  =  180°-  (40° +75°)  =  65°. 

_PR_=_PQ_  RQ        PQ 

sin  Q     sin  R'  sin  P    sin  R' 

/.  P#=-^LsinQ,  /eQ  =  -^%.sinP, 

sin  R  smB 

sin  65°  sin  65° 

*  Results  to  be  written  here. 


SOLUTION  OF  OBLIQUE 

TRIANGLES                   103 

=  .91063X-6428' 

=i&x-9659> 

=  13.24  X. 6428, 

=  13.24  X -9659, 

=  8.51  in. 

=  12.8  in. 

Check:  Take  some  relation  of  Chap.  VI  not  involving  PQ, 
P,  Q,  as  above;  e.g.: 

PR  _  sin  Q 
RQ~ sin  P' 

According  to  this, 

8.51    sin  40°^.  64^8. 

12.8  "  sin  75°     .9659' 

which  gives,  on  multiplying  up, 

8.2198  ...  .  =8.2281. 

This    shows    that   the   results    are    very   nearly    accurate. 
They  are  as  accurate  as  can  be  obtained  with  four-place  tables. 
Another  check:  Ex.  Use  relation  (1),  Art.  40,  as  a  check. 

2.  In  ABC,  vl  =  50o,  B=75°,  c  =  60  in.     Solve  the  triangle. 

3.  In  ABC,  ,4  =  131°  35',  5  =  30°,  6  =  5£  ft.     Find  a. 

4.  In  ABC,  B  =  70°  30',  C=78°  10',  a=102.  Solve  the  tri- 
angle. 

5.  In  ABC,  B  =  9S°  22',  (7=41°  V,  a  =  5.42.  Solve  the  tri- 
angle. 

44.  Case  II.  Given  two  sides  and  an  angle  opposite  to  one 
of  them.  In  the  triangle  ABC  let  a,  b,  A  be  known,  and 
C,  B,  c  be  required.     The  triangle  will  first  be  constructed 


104 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


with  the  given  elements.  At  any  point  A  of  a  straight 
line  LM,  unlimited  in  length,  make  angle  MAC  equal  to 
angle  A,  and  cut  off  AC  equal  to  b.  About  C  as  a  centre, 
and  with  a  radius  equal  to  a,  describe  a  circle.  This  circle 
will  either: 

(1)  Not  reach  to  LM,  as  in  Fig.  88. 

(2)  Just  reach  to  LM,  thus  having  LM  for  a  tangent, 
as  in  Fig.  89. 

(3)  Intersect  LM  in  two  points,  as  in  Figs.  90,  91. 


b 

a 

b, 

0 

0 

A\ 

a 

B 

z> 

c 

a                      / 

/k\ 

V 

a 

'    c~, 

_,_. 

Fig.  88. 


Fig. 


B  M 


B,   --B-'"  B     M 
Fig.  91. 


Each  of  these  possible  cases  must  be  considered.  In 
each  figure,  from  C  draw  CD  at  right  angles  to  AM;  then 
CD  =  b  sin, A. 

In  case  (1),  Fig.  88,  CB<CD,  and  there  is  no  triangle 
which  can  have  the  given  elements.  Hence  the  triangle  is 
impossible  when  a<b  sin  A. 

In  case  (2),  Fig.  89,  CB  =  CD.  Hence,  the  triangle  which 
has  elements  equal  to  the  given  elements  is  right-angled  when 
a  =  b  sin  A. 


SOLUTION  OF  OBLIQUE  TRIANGLES  105 

In  case  (3),  Figs.  90,  91,  CB>CD;  that  is,  a>b  sin  A. 
If  a>b,  then  the  points  B,  Bi,  in  which  the  circle  inter- 
sects LM,  are  on  opposite  sides  of  A,  as  in  Fjg.  90,  and 
there  is  one  triangle  which  has  three  elements  equal  to  the 
given  elements,  namely,  ABC.  If  a<b,  then  the  points 
of  intersection  B,  Bi}  are  on  the  same  side  of  A,  as  in  Fig. 
91,  and  there  are  two  triangles  which  have  elements  equal  to 
the  given  elements,  namely,  ABC,  ABiC.  For,  in  ABC, 
angle  BAC  =  A,  AC=b,  BC=a;  in  ABXC,  angle  B1AC  =  A, 
AC  =  b,  BxC  =  a.  Both  triangles  must  be  solved.  In  this 
case,  Fig.  91,  the  given  angle  is  opposite  to  the  smaller  of 
the  two  given  sides.  Hence,  there  may  be  two  solutions  when 
the  given  angle  is  opposite  to  the  smaller  of  the  two  given 
sides.*  Accordingly  case  II  is  sometimes  called  the  ambigu- 
ous case  in  the  solution  of  triangles. 

Checks:    As  in  Case  I. 

EXAMPLES 
1.  Solve  the  triangle  STV,  given:  ST=15,  VT=12,  S=52°. 

sin  F_sin  S 

sin  V_  sin  52°  =. 7880 
15  12  12   * 

.*.  sin  V=. 9850. 

.-.  7=80°  4',    or    180°-80°4',    i.e.,    99°  56'. 

Both  values  of  V  must  be  taken,  since  the  given  angle  is 
opposite  to  the  smaller  of  the  given  sides.     The  two    triangles 

*  Summary.  When  the  given  angle  is  opposite  to  the  smaller  of  the 
two  given  sides  there  may  be  no  solution  as  in  Fig.  88,  or  one  solution 
as  in  Fig.  89,  or  two  solutions,  as  in  Fig.  91. 

When  the  given  angle  is  opposite  to  the  greater  of  the  two  given  sides 
there  is  one  solution,  as  in  Fig.  90. 


8  V}        V 

Fig.  j2. 


106 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


corresponding  to  the  two  values  of  V  are  STV,  STVh   Fig.  92, 
in  which 

SVT=80°  4';      SVXT=99°  56'. 

InSTVn  In  STV: 

angle  STVi  =  180°-  (S+SVxT)    angle  STV=  180°-  (S+SVT) 


=  28°  4'. 

-  47°, 

56'. 

SVt         VXT 

£7          FT7 

sinSTVx~sinS' 

sin,STK-sinAS" 

SVX        12 
.4705     .7880' 

SV         12 
.7423     .7880' 

/.  #Fi  =  7.165. 

.*.  £7=11.3. 

The  solutions  are: 

Fi  =  99°56'  ' 

7=80°    4'  ' 

STVi  =  28°    4' 

STV =47°  56' 

■  . 

aSFi  =  7.165     J 

£7=11.3       j 

Check:  Of  several  possible  checks  use  relations  (1),  Art.  40. 
InSTVf.  In  STV: 


ViT-SVi    Umt(SSTVt) 


V^+SVi     tani(S+STVi)' 

On  substituting  the  values  above  there  comes, 


VT-SV    tan  j(S- STV) 
VT+SV~~  tan  ^(S+STV)' 


4.835  _  tan  11°  58' 
19. 165 ""tan  40°    2'* 

4.835      .2100 


.7  =tan    2°    2' 
23.3 "tan  49°  58'" 

.7        .0355 


19.165     .8401'  23.3     1.1904' 

From  these,  on  clearing  fractions, 

4.0619=4.0246,  .83328=  .82715. 

N.B.  Had  five-place  or  six-place  tables  been  used,  the  cal- 


SOLUTION  OF  OBLIQUE  TRIANGLES  107 

culated  values  would  have  approximated  still  more  closely  to- 
absolute  correctness. 

This  shows  that  the  values  found   very  nearly  satisfy  the 
check,  and  accordingly  closely  approximate  to  correctness. 

2.  Solve  ABC,  given:  a  =  29  ft.,  6  =  34  ft.,  4  =  30°  20'. 

3.  Solve  ABC  when     a  =  30  ft.,  6  =  24  ft.,  5  =  65°. 

4.  Solve  ABC  when     a  =  30  in.,  b  =  24  in.,  A  =  65°. 

6.  Solve  ABC  when     a=  15  ft.,  b  =  8  ft.,  B=2S°  25'. 

45.  Case  III.     Given  two  sides  and  their  included  angle. 

In  the  triangle  ABC,  a,  b,  C,  say,  are  known,  and  it  is  required 
to  find  A,  B,  c.  In  this  case,  c  can  be  deter- 
mined from  the  relation 

c2  =  a2  +  b2— 2ab  cos  C; 

angle  A  can  be  determined  from  the  relation 

A        c        b  sin  A     sin  C 

Fig.  93.  ~1T  =  ~C~; 

angle  B  can  be  determined  from  the  relation 

~     „  r,       ,        sin  B    sin  C 

A+B  +  C  =  1S0,°  or  from  — ,—  = . 

o  c 

Checks:  Any  relations  in  Chap.  VI  which  have  not  been 
used  in  the  above  solution. 

EXAMPLES 
1.  In  triangle  PQR,  p  =  8ft.,  r=10ft.,   Q  =  47°.     Find  q, 

q2==p2_^_r2_2pr  cos  Q 

=  64  + 100  -  2  X  8  X 10  X  .6820  =  54.88. 
■■  ?=7.408.  '   FlG.94.    « 


108  ELEMENTS  OF  PLANE- TRIGONOMETRY 

sin  p=£sinO==8X7|4=  ^  p=520  ^ 

g  7.408 

sin  R=:A^R  =  10X_7314  =  .9873.  ...  7^80°  50'. 

5  7.408 

Cfadfc:  Of  the  available  checks  use  P+Q  +  R=180°. 
Here  52°  10' +47° +80°  50'=  180°; 

and  the  test  is  thus  satisfied. 

2.  Solve  ii£C,  given:  a=34  ft.,  6  =  24  ft.,  C=59°  17'. 

3.  Solve  45C,  given:  a  =  33  ft.,  c  =  30  ft.,  5  =  35°  25'. 

4.  Solve  RST,  given:  r=30  ft.,  s  =  54  ft.,  77=46°. 

5.  Solve  PQ#,  given:  p=  10  in.,  5=  16  in.,  #  =  97°  54'. 

46.  Case  IV.  Three  sides  given.  If  the  sides  a,  b,  c  are 
known  in  the  triangle  ABC,  then  the  angles  A,  B,  C  can 
be  found  by  means  of  the  relations  (2),  Art.  39,  or  the 
relations  in  Art.  41. 

Checks:  Any  relations  in  Chap.  VI  which  have  not  been 
used  in  the  solution. 

EXAMPLES 

1.  In  ABC,  a=±,  6  =  7,  c=10;  find  A,  B,  C. 

&+c2-a2    49+100-16      133       n_nn     .      .      ieolo, 
COsA  =  -2bc—=    2X7X10    =l4Q-=-950Q-   "  ^  =  18°12' 


D     c2  +  a2-V>     100+16-49       67        0<VTC 
C0*B=—2^ =    2X10X4    =^0-=-8375- 


£=33°    7'.5. 


a2  +  b2_c2     16+49-100     -35  _0_n 

CQS  C7=— 2^6-  =     2X4X7     =-5^=--6250' 

.'.  C=128°40'.8. 


SOLUTION  OF  OBLIQUE  TRIANGLES  109 

Angle  C  is  in  the  second  quadrant  since  its  cosine  is  negative. 
Check:   18°  12'  +  33°  7'.5  +  128°  40'.8=180° 
JS^     0'.3.     The  discrepancy  is  due  to  the  fact  that 
four-place  tables  were  used  in  the  computation. 
Had    five-place    tables    been    used,    the    dis- 
crepancy would  have  been  less. 

2.  In  PQR,  p=  9,  9  =  24,  r=27.     Find  P,  Q,  R. 

3.  In  RST,  r=21,  s  =  24,   t  =  27.     Find  R,  S.  T. 

4.  In  ABC,  a  =12,  6  =  20,   c  =  28.     Find  A,  B,  C. 

5.  In  ABC,  a  =  80,  6  =  26,  c  =  74.     Find,  A,  B,  C. 

6.  Solve  Ex.  1,  using  five-place  tables. 

7.  Solve  several  of  Exs.  1-5,  using  relations  in  Art.  41. 

47.  Use  of  logarithms  in  the  solution  of  triangles.  If 
logarithms  are  not  employed  all  the  relations  in  Arts.  38-41 
are  available  for  the  solution  of  triangles.  If  logarithms 
are  employed,  only  those  relations  which  are  adapted  to 
logarithmic  computation  can  be  used,  viz.,  the  relations  in 
Arts.  38,  40,  41.  The  relations  in  Art.  39  are  not  adapted 
for  logarithmic  computation. 

The  examples  worked  in  Arts.  48-50  will  give  sufficient 
explanation  of  how  logarithms  are  used  in  the  solution  of 
triangles. 

48.  Cases,  I,  II,  logarithms  used. 

EXAMPLES 
1.  In  ABC,  given:  a =447,     To  find:  5=  (Write the  results 
6=576,  C=         here.) 

A =47°  35'.  s     c= 


110 


ELEMENTS  OF  PLANE  TRIGONOMETRY 


Since  a<b,  there  may  be  two   solutions. 
Construction  shows  there  are  two  solutions. 

Formulas:    sin  A BC=  —  sin  A  =  sin  AB\C. 
a 

ACB  =  180°-  (A  +  ABC). 

ACBX  =  180°-  (A  +  ABlC). 

a 


A        c  Bx 
Fig.  96. 


AB= 


sin  ACB. 


AB^-^-r  sin  ACBh 


sin  A  "'  ~  *     sin  A 

.'.  log  sin  ABC =\og  6  + log  sin  A  — log  a  =  log  sin  ABiC; 
log  AB  =  log  a  +  log  sin  ACB  —  log  sin  A; 
log  AB i  =  log  a+log  sin  ACB\  —  log  sin  A. 
log  a  =  2.65031 
log  6=2.76042 
log  sin  A  =  9.86821  -10 
.-.  log  sin  5= 9.97832- 10 


.-.  ABC  =72°  2'  45" 
.-.  ACB  =  60°  22'  15'' 
log  sin  ACB  =  9.93914- 10 
/.  logA£  =  2.72124 
.'.  AB  =  o2G.3 


and  ABXC=  107°  57'  15" 
.'.  AC£i  =  24°27'45" 
log  sin  ACBX  =  9.61710- 10 
/.  log  A Bx  =  2.39920 
.*.  AB1  =  250.7 


In  obtaining  log  A B,  for  instance,  log  sin  ACS  may  be 
written  on  the  margin  of  a  slip  of  paper,  placed  under  log  a,  the 
addition  made,  log  sin  A  placed  beneath,  and  the  subtraction 
made. 

Solve  the  triangle  ABC  when  the  following  elements  are 
given,  and  check  the  results: 

2.  A  =  63°  48',  £=49°  25',  a  =  825  ft. 


SOLUTION  OF  OBLIQUE  TRIANGLES 


111 


3.  B=  128°  3'  49",  (7=33°  34' 47",  a=240ft. 

4.  4  =  78°  30',  6=137  ft.,  a  =  65  ft. 

5.  a=275.48,  6  =  350.55,  5=60°  0'  32". 

6.  c= 690,  a =464,  4  =  37°  20'. 

7.  a =690,  6=1390,  A  =  21°  14'  25". 
49.  Case  III,  logarithms  used. 

EXAMPLES 

1.  In  triangle  ABC,  given:  6  =  472, 

c  =  324, 

,4  =  78°  40'. 

b  —  c 
Formulas:     tan  h(B—C)=T- —  cot  \A 

b  +  c 


£=«£+C)+K5-C). 

C=iCB  +  C0-H£-C). 


Fig.  97. 


6  sin  A 


or   =  ■ 


sin  ^4 


sin  B  '  sin  C  ' 

logtani(^-Q=log(6-c)+logcotJA-log(6  +  c), 
log  a  =  log  6  +  log  sin  A  —  log  sin  B ;     or 
=  log  c  +  log  sin  A  —  log  sin  C. 
log  (b-c)  =  2.17026  log  6=2.67394 

log  (b+c)  =  2.90091  log  sin  A  =9.99145-10 

log  cot  \A  =10.08647-10  log  sin  5  =  9.95160-10 


6=472 
c=324 
^=78°  40' 
6-c  =  148     / 
6+c=796 
\A  =39°  20' 


logtan*(B-C)=  9.35582-19 
/.  *(£-C)=12°47'  1" 
i(£+C)=50°40' 
.*.  5=63°  27'  1" 
.*.  C=37°  52'  59'! 


log  a  =  2.71379 
.'.  a=517.36 


112  ELEMENTS   OF  PLANE  TRIGONOMETRY 

Check:  a=c  sin  A-r-sin  C. 
Solve  the  following  triangles  and  check  the  results: 

2.  ABC,  given  6  =  352,  a=266,  C=73°. 

3.  PQR,  given  p=*91.7,  q= 3 1.2,  #  =  33°  7'  9", 

4.  A£C,  given  a=960,  6  =  720,  C=25°  40'. 

'6.  4£C,  given  6  =  9.081,  c=3.6545,  4  =  68°  14'  24". 
50.  Case  IV,  logarithms  used. 

EXAMPLES 

1.  In   triangle   ABC,    a  =  25. 17,    6  =  34.06, 
c-22.17.     Find  A,  B,  C. 


FormvUs:        r=  yj(*~  «)(*-»(*-*) 


s 


A    C-  22.17  £ 


r  r  r 

tan£^.= ;     tan  ££  = r;     tan  4(7= .  Fig.  98 

s— a  s— o  s—c 

.'.  logr=£[log  (s-a) +log  (s-6)+log  (s  —  c) -logs]. 

log  tan  £4  =  log  r-log  (s-a);       log  tan  ££  =  log  r-log  (s-6); 

log  tan  £C=log  r— log  (s—c). 

Check:  ^+£+0=180°. 

a=25.17  log  s=  1.60959      log  tan  £4=   9.64465-10 

6  =  34.06      log  (s-a)  =  1.19117  ^4  =  23°  48'  28" 

c=22.17      log  (s-6)  =0.82217      log  tan  £5=10.01365-10 


2s=81.40      log  (s-c)  =  1.26788  £5=45°  54 


s=40.70  logtan£C=   9.56794-10 


SOLUTION  OF  OBLIQUE  TRIANGLES  113 

s-a=  15.53         .*.  log  r2=  1.67163  £C=20°  17'  35" 

s-b=   6.64  .'.  logr=0.83582 

s-c=  18.53         .*.  4  =  47°  36' 56",  5=91°  48',  C=40°35'  10" 

Check:  A+B+C=180°  0'  6". 

Solve  the  following  triangles  and  check  the  results: 

2.  ABC,  given  a  =  260,  6  =  280,  c=300. 

3.  ABC,  when  a  =  26.19,  6  =  28.31,  c  =  46.92. 

4.  PQR,  given  p  =  650,  g=736,  r=914. 

5.  RST,  given  r=1152,  s=2016,  Z=2592. 

51.  Problems  in  heights  and  distances.  Some  problems  in 
heights  and  distances  have  been  solved  in  Art.  11  by  the 
aid  of  right-angled  triangles.  Additional  problems  of  the 
same  kind  will  now  be  given,  in  the  solution  of  which  oblique- 
angled  triangles  may  be  used.  It  is  advisable  to  draw  the 
figures  neatly  and  accurately.  The  graphical  method  should 
also  be  employed. 

EXAMPLES 

1.  Another  solution  of  Ex.  2,  Art.  11. 

In  the  triangle  ABP  (Fig.  23),  45=100  ft.,  BAP=30°, 
PBA  =  1S0°— 45°=  135°.  Hence  the  triangle  can  be  solved, 
and  BP  can  be  found.  When  BP  shall  have  been  found,  then 
in  the  triangle  CBP,  BP  is  known  and  CBP= 45°;  hence  CP 
can  be  found.     The  computation  is  left  to  the  student. 

2.  Another  solution  of  Ex.  3,  Art.  11.  In  the  triangle 
CBP  (Fig.  24),  £P=30ft.,  BCP= 40°  20' -38°  20' =2°, 
PBC  =  90°  +  LCB=  128°  20'.  Hence  CBP  can  be  solved  and 
the  length  of  CB  can  be  found.  When  CB  shall  have  been 
found,    then,   in   the  triangle  LCB,  angle  C=38°  20',    CB  is 


114  ELEMENTS  OF   PLANE   TRIGONOMETRY 

known,  and  hence  LB  can  be  found.    The  computation  is  left 
to  the  student. 

3.  Find  the  distance  between  two  objects  that  are  invisible 
from  each  other  on  account  of  a  wood,  their  distances  from  a 
station  at  which  they  are  visible  being  441  and  504  yd.,  and  the 
angle  at  the  station  subtended  by  the  distance  of  the  objects 
being  55°  40'. 

4.  The  distance  of  a  station  from  two  objects  situated  at 
opposite  sides  of  a  hill  is  1128  and  936  yd.,  and  the  angle 
subtended  at  the  station  by  their  distance,  is  64°  28'.  What 
is  their  distance? 

5.  Find  the  distance  between  a  tree  and  a  house  on  oppo- 
site sides  of  a  river,  a  base  of  330  yd.  being  measured  from 
the  tree  to  another  station,  and  the  angles  at  the  tree  and 
the  station  formed  by  the  base  line  and  lines  in  the  direction 
of  the  house  being  73°  15'  and  6S°  2',  respectively.  Also  find 
the  distance  between  the  station  and  the  house. 

6.  Find  the  height  of  a  tower  on  the  opposite  side  of  a 
river,  when  a  horizontal  line  in  the  same  level  with  the  base 
and  in  the  same  vertical  plane  with  the  top  is  measured  and 
found  to  be  170  ft.,  and  the  angles  of  elevation  of  the  top  of 
the  tower  at  the  extremities  of  the  line  are  32°  and  58°,  the 
height  of  the  observer's  eye  being  5  ft. 

7.  Find  the  height  of  a  tower  on  top  of  a  hill,  when  a  hori- 
zontal base  line  on  a  level  with  the  foot  of  the  hill  and  in 
the  same  vertical  plane  with  the  top  of  the  tower  is  measured 
and  found  to  be  4G0  ft.;  and  at  the  end  of  the  line  nearer 
the  hill  the  angles  of  elevation  of  the  top  and  foot  of  the 
tower  are  36°  24',  24°  36',  and  at  the  other  end  the  angle  of 
elevation  of  the  top  of  the  tower  is  16°  40'. 

8.  A  church  is  at  the  top  of  a  straight  street  having  an 
inclination  of  14°  10'  to  the  horizon;  a  straight  line  100  ft. 
in  length  is  measured  along  the  street  in  the  direction  of  the 


SOLUTION  OF   OBLIQUE  TRIANGLES  115 

church;  at  the  extremities  of  this  line  the  angles  of  elevation 
of  the  top  of  the  steeple  are  40°  30',  58°  20'.  Find  the  height 
of  the  steeple. 

9.  The  distance  between  the  houses  C,  D,  on  the  right 
bank  of  a  river  and  invisible  from  each  other,  is  required. 
A  straight  line  AB,  300  yd.  long,  is  measured  on  the  left  bank 
of  the  river,  and  angular  measurements  are  taken  as  follows: 
ABC=53°  30',  CBD  =  Vo°  15',  CAD  =  S7°,  DAB=58°  20'. 
What  is  the  length  CD? 

10.  A  tower  CD,  C  being  the  base,  stands  in  a  horizontal 
plane;  a  horizontal  line  AB  on  the  same  level  with  the  base 
is  measured  and  found  to  be  468  ft. ;  the  horizontal  angles 
BAC,  ABC,  are  equal  to  125°  40',  12°  35',  respectively,  and 
the  vertical  angles  CAD,  CBD,  are  equal  to  38°  20',  11°  50', 
respectively.  Find  the  height  of  the  tower  and  its  distances 
from  A  and  B. 

11.  A  base  line  AB  850  ft.  long  is  measured  along  the 
straight  bank  of  a  river;  C  is  an  object  on  the  opposite  bank; 
the  angles  BAC,  ABC,  are  observed  to  be  63°  40',  37°  15', 
respectively.     Find  the  breadth  of  the  river. 

12.  A  tower  subtends  an  angle  a  at  a  point  on  the  same 
level  as  the  foot  of  the  tower  and,  at  a  second  point,  h  feet 
above  the  first,  the  depression  of  the  foot  of  the  tower  is  /?. 
Show  that  the  height  of  the  tower  is  h  tan  a  cot  /?. 

13.  The  elevation  of  a  steeple  at  a  place  due  south  of  it 
is  45°;  at  another  place  due  west  of  the  steeple  the  elevation 
is  15°.  If  the  distance  between  the  two  places  be  a,  prove 
that  the  height  of  the  steeple  is 

a(v/3-l)-2v/2. 

14.  The  elevation  of  a  steeple  at  a  place  due  south  of  it 
is  45°;     at  another  place  due  west    of    the    first    place    the 


116  ELEMENTS  OF  PLANE  TRIGONOMETRY 

elevation   is    15°.     If  the   distance  between   the   two   places 
be  a,  prove  that  the  height  of  the  steeple  is 

a(V3-  l)-s-2  a/3. 

15.  The  elevation  of  the  summit  of  a  hill  from  a  station  A 
is  a ;  after  walking  c  feet  toward  the  summit  up  a  slope  inclined 
at  an  angle  /?  to  the  horizon  the  elevation  is  y.  Show  that 
the  height  of  the  hill  above  A  is  c  sin  a  sin  iy—ft)  cosec  (y— a)  ft. 


CHAPTER  VIII 


MISCELLANEOUS  THEOREMS 

52.  Area  of  a  triangle.     The  area  of  ABC  is  required. 
Let  the  length  of  the  perpendicular  DC  from  C  to  AB,  or 


A  D      B 

Fig.  99. 

AB  produced,  be  denoted  by  p,  the  semi-perimeter  by  s, 
and  the  area  by  S.    The  following  cases  may  occur: 

I.  One  side  and  the  perpendicular  on  it  from  the  opposite 
angle  known,  say  (c,  p). 

S=icp.  (1) 

II.  Two  sides  and  their  included  angle  known,  say,  b,  c,  A. 

S  =  icp  =  %c.  AC  sin  BAC. 


S  —  \bc  sin  A. 


(2) 


Simikrly ,        S  =  \ca  sin  i?  =  \ab  sin  C. 

Problems  which  do  not  fall  under  Cases  I  or  II  or  III 
directly  may  be  solved  on  finding  a  perpendicular  or  a  side 
or  an  angle. 


117 


118  ELEMENTS    OF    PLANE    TRIGONOMETRY 

Eg.  Let  a,  A,  B,  C,  be  known. 

a  sin  B 


Now  S  =  %ab  sin  C.     But  6  = 
.*.  S=ia2 


sin  A 
sin  C  sin  B 


smA 

III.  Three  sides  known. 

S  =  \bc  sin  A  =  \bc-2  sin  \A  cos  }A 


=  fo    /(s-ft)Q-c)    js(s-a) 
\  6c  \      be     ' 

.-.    S  =  V8(8-a)(8-b){8-c).  (3) 

EXAMPLES 

1.  Find  the  area  of  the  following  triangles : 

(a)  A£C  in  which  a=30  ft.,  6  =  36  ft.,  c  =  44  ft. 
(6)   PQi2  in  which  p  =  22  ft.,  g=31  ft.,  r=43  ft. 

2.  Find  the  area  of  the  following  triangles : 

(a)  ABC  in  which  a  =  37  ft.,  6  =  53  ft.,  C=43°. 
(6)   PQR  in  which  g=23  ft.,  r=48  ft.,  P=65°. 

3.  An  isosceles  triangle  whose  vertical  angle  is  78°  contains 
400  sq.yd.;  find  the  lengths  of  the  sides. 

4.  Find  two  triangles  each  of  which  has  sides  63  and  55  ft. 
long,  and  an  area  of  874  sq.ft. 

5.  Two  roads  form  an  angle  of  27°  10r.  At  what  distance 
from  their  intersection  must  a  fence  at  right  angles  to  one 
of  them  be  placed  so  as  to  inclose  an  acre  of  land? 

6.  The  angles  at  the  base  of  a  triangle  are  22°  30'  and  112° 
30',  respectively:  show  that  the  area  of  the  triangle  is  equal 
to  the  square  of  half  the  base. 

53.  Area  of  a  circular  sector.     Let  r  be  the  radius  of  the 


MISCELLANEOUS  THEOREMS  119 

circle  and  0  the  number  of  radians  in  the  angle  of  the  sector 
j?       AOB.    Now       . 

area  of  sector  _   angle  AOB        d  radians 
area  of  circle  ~4  right  angles  ™"  2n  radians' 


area  of  sector 
i.e.. 


nr2  2n 

.•.    area  of  sector=l^2e. 

m 

This  formula  is  not  true  unless  the  angle  is  measured  in 
radians. 

Otherwise:     Area  sector  =  \r  •  arc  AB. 
Now  arcAB  =  rd.  [Art.  18  (5).] 

.*.  area  sector  =  \r2d. 

EXAMPLES 

1.  Draw  the  following  sectors  and  calculate  their  arcs  and 
areas : 

(a)  Radius  =10  in.,  angle  =  f  radian. 

(b)  Radius  =  24  in.,  angle  =1J  radians. 

(c)  Radius  =18  in.,  angle  =  2  radians. 

(d)  Radius  =  20  in.,  angle  =  4J  radians. 

2.  The  arc  of  a  sector  =  24  in.  and  its  angle  =  J  radian: 
find  the  radius  and  the  area  of  the  sector. 

3.  The  area  of  a  sector  is  124  sq.in.  and  its  angle  is  2  radians: 
find  the  lengths  of  its  radius  and  arc. 

4.  The  area  of  a  sector  is  236  sq.in.  and  its  arc  =  32  in.: 
find  the  radius  and  the  angle  of  the  sector  in  radians  and  in 
degrees. 

54.  Circles  connected  with  a  triangle.  A.  The  circum- 
scribing circle.     Let  S  denote  the  area  of  the  triangle  ABC 


120 


ELEMENTS  OF   PLANE  TRIGONOMETRY 


and  R  the  radius  of  its   circumscribing  circle.     Then  by- 
Art.  38  (3),  (4), 


2S  = 


2sin^     2  sin  2*     2sinC' 


(1) 


2S 
From  (2),  Art.  52,  sin  A  =  ^. 

first  of  equations  (1),  gives 

abc 

4S' 


Substitution  of  this  in  the 


U  = 


(2) 


B.  The  inscribed  circle.     Let  the 

radius  of  the  circle  inscribed  in  a  tri- 
angle ABC  be  denoted  by  r.  Join 
the  centre  0  and  the  points  of  contact 
L,  M,  N.  By  geometry,  the  angles 
at  L,  M,  N  are  right  angles.  Draw  A 
OA,  OB,  OC. 

Area  £OC  +  area  CO  A 

+area  AOB  =  area  ABC. 


Fig.  102. 


i.e. 


iar  +  ibr  +  icr  =  Vs(s— a)(s—  b)(s— c),  or  S. 
;.  b(a  +  b+c)r  =  S, 


.-.    r- ^ 


s-a)(s-b)(s-c)    8 


(3) 


This  was  shown  in  another  way  in  Art.  41. 

C.  The  escribed  circle.     An  escribed  circle  of  a  triangle  is 


MISCELLANEOUS  THEOREMS 


121 


a  circle  that  touches  one  of  the  sides  of  the  triangle  and 

the  other  two  sides  produced. 

Let  ra  denote  the  radius  of  the 
escribed  circle  touching  the  side  BC 
opposite  to  the  angle  A.  Join  the 
centre  Q  and  the  points  of  contact 
L,  M,  N.  By  geometry,  the  angles 
at  L,  M,  N  are  right  angles.  Draw 
QA,  QB,  QC. 

Area  ABQ +area  CAQ— area  BCQ  =  area  ABC, 

.'.  Jrac  +  Jra6— iraa  =  S, 

.'.  %(c  +  b  —  a)ra=S; 

(s—a)ra=S. 

S 

.\    ra  = . 

s-a 


Similarly, 


S 


s 


rb  = 


s-b1 


rr  = 


s—c 


EXAMPLES 

1.  Find    the    radii    of    the    inscribed,    circumscribed,    and 
escribed  circles  of  the  following  triangles: 

(a)  ABC  in  which  a  =  22  in.,  6  =  35  in.,  c=43  in. 

(6)   PQR  in  which  p=  10  in.,  q=  13  in.,  r=   8  in. 

(c)    RST  in  which  r=32  in.,  s  =  40  in.,  £  =  50  in. 

2.  Prove  that  in  an  equilateral  triangle  the  radii  of  the 
inscribed,  circumscribed,  and  escribed  circles  are  as  1:2:3. 

3.  The  sides  of  a  triangle  are  17  in.,  25  in.,  36  in.;    show 
that  the  radii  of  the  escribed  circles  are  as  21 :33: 154. 


122  ELEMENTS  OF  PLANE  TRIGONOMETRY 

4.  Prove:  (a)  ra  +  rb  +  rc—r=4:R;       (6)   Vr-ra-rb-rc  =  S. 

5,  Prove:  (a)  I-pJr+i--;  (&)  #r= 


ra     r6     rc     r'  4(a  +  6  +  c)' 

(c)  r=4/c  sin  —  sin  —  sm  — . 

Z  Z  Z 

6.  Prove  ra  cot  —  =  rb  cot  —  =  rc  cot  —  =  r  cot  —  cot  —  cot  — 

z  z  z  z        z        z 

AT?  A  B  C 

=  4/c  cos  —  cos  —  cos  — . 

z         z        z 

a  A  A 

7.  Prove  R  =  —. — j,  r=(s  —  a)  tan—,  ra  =  s  tan—.     Write 

z  sm  A.  z  z 

two  other  similar  formulas  for  R  and  r.     Write  similar  formulas 
for  rb  and  rc. 

.    B  .    C 

a  sm  —  sm  — 

8.  Prove    r= -. .     Write    two    similar  formulas 


involving  b,  c. 


A 
cos2 


9.  (a)  Show  that  the  area  of  a  regular  polygon  inscribed 
in  a  circle  is  a  mean  proportional  between  the  areas  of  an 
inscribed  and  circumscribing  polygon  of  half  the  number  of 
sides,  (b)  The  sides  of  a  triangle  are  as  2:3:4;  show  that  the 
radii  of  the  escribed  circles  are  as  J:J:1. 

10.  If  the  altitude  of  an  isosceles  triangle  is  equal  to  its 
base,  the  radius  of  the  circumscribing  circle  is  f  of  the  base. 

11.  An  equilateral  triangle  and  a  regular  hexagon  have  the 
Same  perimeter.  Show  that  the  areas  of  their  inscribed 
circles  are  as  4 : 9. 

12.  If  the  sides  of  a  triangle  are  51,  68,  and  85  ft.,  show 
that  the  shortest  side  is  divided  by  the  point  of  contact  of  the 
inscribed  circle  into  two  segments,  one  of  which  is  double  the 
ether. 


MISCELLANEOUS  THEOREMS 


123 


55.  Relations  between  the  radian  measure,  the  sine,  and 
the  tangent  of  certain  angles. 

A.  If  0  be  the  radian  measure  of  an  acute  angle, 

sin  0<8<tan  0. 

B.  When  an  angle  0  radians  approaches  the  limit  zero, 

sin  0  •       tan  0 

~T    and     ~~e~ 

each  approaches  unity  as  a  limit. 

Proof  of  A.     Let  Angle  AOP  =  0  radians. 

Make  angle  AOR  =  AOP.    With  radius  r  describe  the 
p  arc     QBR    about    0    as    centre. 

Draw  the  chord  QR  intersecting 
OA  at  M.  Draw  the  tangents  QT, 
RT,  intersecting  OA  at  T.  Draw 
the  chord  QB*    Then 

iB  ,/T  A  area  triangle  OQB  <  area  sec  tor  OQB 
I  / 

<  area  triangle  OQT. 

-Vjt  That  is; 

\0B  •  MQ  <  iOQ  •  arc  BQ  <  %0Q  •  QT; 


o\ 


M 


Fig.  104 


or, 


Jr-r  sin  0<%r-rO<%r-r  tan  'd. 
sin  0  <  0  <  tan  0. 


Hence, 

Otherwise:  It  can  be  proved  that 

QMR<sltcQBR<QTR. 
From  this,  on  dividing  by  2, 


(1) 


*  Or  suppose  it  is  drawn. 


124  ELEMENTS  OF  PLANE  TRIGONOMETRY 

MQ  <  arc  QB<QT. 

That  is,  r  sin  0  <      rd     <  r  tan  0. 

.'.  sin  0<       d      <tan  6. 

Proof  of  B.     Division  of  each  member  of  (1)  by  sin  6  gives 

6  1 

l<ir_-^< 


sin  6    cos  0' 

a 

Thus,  for  0  between  0°  and  90°,  - — ^  lies  between  1  and 
• -q.    Now  when  6  approaches  zero,  cos  d  approaches  1  as 

a  limit  and  thus 7  approaches  1  as  a  limit.    Accordingly 

o 

the  limit  of  - — 7,  which  must  lie  between  1   and   the  limit 
sin  0' 

of ™  viz.,  1,  is  itself  1.     Hence  the  limit  of  the  reciprocal 

sin  6  .    „ 

-y-isl. 

Division  of  each  member  of  (1)  by  tan  0  gives 

0 

cos  d<- B<1. 

tan  0 

a 

Thus,  for  0  between  0°  and  90°,  ? 7  lies  between  cos  6 

'  '  tan  d 

and  1.    Now  when  6  approaches  zero,  cos  6  approaches  1  as 

a  limit.    It  follows,  on  reasoning  as  in  the  preceding- case, 

0  ,  ,    tan  0  •     , 

that  7 — n,  and  consequently  — 7—,  approaches  1  as  a  limit. 

These  results  may  be  briefly  expressed : 


Limit /sin  9\  Limit /tan  0\ 

e=oV~e~7  =  1;     e=o\-r"/=1- 


(2) 


MISCELLANEOUS  THEOREMS  125 

These  are  two  of  the  most  important  theorems  in  ele- 
mentary trigonometry;  they  are  frequently  employed  both 
in  practical  work  and  in  pure  mathematics. 

A  very  important  corollary  to  (2)  is  the  following: 
If  d  be  the  radian  measure  of  a  very  small  angle,  then  6  can 
be  used  for  sin  0  and  tan  d  in  calculations. 

For  instance,  sin  10"  to  12  places  of  decimals  is 
.000048481368.  This  is  also  the  radian  measure  of  10" 
to  12  places  of  decimals.  The  radian  measures,  sines,  and 
tangents,  of  angles  from  0°  to  6°,  agree  in  the  first  three 
places  of  decimals.     For 

radian  measure  6°  =  (.  10472)  = .  105 ;     sin  6°  =  (.  10453)  -  .105; 

tan  6°=  (.10510)  =  .105. 

EXAMPLES 

1.  Find  the  angle  subtended  by  a  man  6  ft.  high  at  a  dis- 
tance of  half  a  mile. 

-H= '"*       Here'   ^tan^  =  2olo  =  4i0- 

Fig.  105.  Now 

V  _    1       18Q°_  7X108°  , 

440     440        7i       22X440 

2.  What  must  be  the  height  of  a  tower,  in  order  that  it 
subtend  an  angle  1°  at  a  distance  of  4000  ft.? 

•=tan  1°  =  radian  measure  1°=— -- 


4000  180 

22 


4000 


7X180'  Fig.  106. 


.    x     22X40Q0-60£lft 
••  X~   7X180    -69-84tt- 

3c  Verify  the  following  statements: 


126  ELEMENTS  OF  PLA1<TE  TRIGONOMETRY 

An  angle  1°  is  subtended  by  1  in.  at  a  distance  4  ft.  9.3  in., 
and  by  1  ft.  at  a  distance  57.3  ft.  An  angle  V  is  subtended 
by  1  in.  at  a  distance  286.5  ft.,  and  by  1  ft.  at  a  distance 
3437.6  ft.,  about  two-thirds  of  a  mile.  An  angle  1"  is  sub- 
tended by  1  in.  at  a  distance  of  nearly  3J  mi.,  by  1  ft.  at  a 
distance  a  little  greater  than  39  mi.,  by  a  horizontal  line  200  ft. 
long  on  the  other  side  of  the  world,  nearly  8000  mi.  away. 

4.  The  moon's  mean  angular  dfameter  as  observed  at  the 
earth  is  31'  5",  and  its  actual  diameter  is  about  2160  miles. 
Find  the  mean  distance  of  the  moon.  How  many  full  moons 
would  make  a  chaplet  across  the  sky? 

5.  Taking  the  earth's  equatorial  radius  as  3963  mi.,  find  the 
angular  semi-diameter  of  the  earth  as  it  would  appear  if 
observed  from  the  moon.  Compare  the  relative  apparent  sizes 
of  the  moon  as  seen  from  the  earth,  and  the  earth  as  seen 
from  the  moon. 

6.  The  semi-diameter  of  the  earth  as  seen  from  the  sun  is 
very  nearly  8".8.  What  is  the  sun's  distance  from  the  earth, 
the  radius  of  the  earth  being  assumed  as  4000  miles? 

7.  At  least  how  many  times  farther  away  than  the  sun 
is  the  nearest  fixed  star  a  Centauri,  at  which  the  mean  distance 
between  the  earth  and  the  sun  (about  92,897,000  miles)  sub 
tends  an  angle  something  less  than  1"?     How  long,  at  least, 
will  it  take  light  to  come  from  this  star  to  the  earth? 

8.  Find  approximately  the  distance  at  which  a  coin  an  inch 
in  diameter  must  be  placed  so  as  just  to  hide  the  moon,  the 
latter's  angular  diameter  being  taken  31'  5". 

9.  The  inclination  of  a  railway  to  a  horizontal  plane  is  50'. 
Find  how  many  feet  it  rises  in  a  mile. 


MISCELLANEOUS  THEOREMS  127 

10.  Find  the  angle  subtended  by  a  circular  target  4  ft.  in 
diameter  at  a  distance  of  1000  yd. 

11.  Find  the  height  of  an  object  whose  angle  of  elevation 
at  a  distance  of  900  yd.  is  1°. 

12.  Find  the  angle  subtended  by  a  pole  20  ft.  high  at  a 
distance  of  a  mile. 


ANSWERS 


Art.  3,  Pages  4,  5 

2.  .3025,  .9732,  .9502,  .4754,  5.4623,  .3167,  9.78280,  9.50336,  10.73227. 

3.  8°  48'.8,  38°  26'.7, 
30°  43'  ,  52°  3'.2, 
23°    0\9,              31°22/.5, 

28°  18'  22".5, 
22°  20'  30", 
33°  31'  31".l. 

Art.  4,  Pages  7,  8 

In  the  following  examples  the  functions  are  given  in  the  order  on 
page  3. 

5.  .8575,  .5145,  1.6667,  .6,  1.9437,  1.1662,  59°  2'.2. 

6.  .4,  .9165,  .4364,  2.2913,  1.0911,  2.5,  23°  35';    .3162,  .9487,  .3333,  3, 

1.0541,  3.1623,  18°  25'.9. 

7.  9035,  .4286,  2.1082,  .4743,  2.3333,  1.1068,  64°  37'.3. 

11.  (1)  a:b,     Vb2-a2:b,     a:Vb2-a\     Vb2-a2:a,     b:Vb2-a2,  b:a; 

(2)  Vb2-a2:b,    a:b,    Vtf-tf-.a,    a:Vb2-a2,    b:a,    b:Vb2~a2; 

(3)  a:\/a2  +  &2,    b:Va2+b2,    a:b,    b:a,    Va2+b2:b,    Va2+b2:a; 
(4)6:Va2  +  &2,    a:Va2+62,    b:a,    a:b,    Va2  +  b2:a,    Va2  +  b2:b; 

(5)  Va2-b2:a,    b:a,    Va2-b2:b,    b:Va2-b2,    a:b,    a:Va2-b2; 

(6)  b\a,  xftf-W'.a,  biVtf^b2,  Vtf^Wib,  a:Va2-b\  a:b. 

12.  41°  24'  35". 

13.  19°  28'  16". 

Art.  5,  Page  11 

1.  2.28025.  4.  2.0404.  7.  5.9259. 

2.  2.3333.  5.  2.25.  8.  .3248. 

3.  5.846.  6.  4.619.  9.  2.75. 

10.   -.708. 

128 


ANSWERS 


129 


Art.  6,  Pages  12,  13 

2.  cos  11°  40',  sin  9°  30',  cot  40°,  tan  25°,  cosec  19°,  sec  10c 

3.  16°  40'.  4.  40°. 

Q  Art.  7,  Pages  1G,  17 


sin  A 

cos  A 

tan  A 

cot  A 

sec  A 

cosec  A 

sin  A 

tan  A 

1 

VSec2A-l 

1 

sin  A 

Vl-cos2A 

Vi  +  tan2^ 

Vi+Cot2A 

sec  A 

cosec  A 

cos  A 

1 

cot  A 

1 

sec  A 

V  cosec2  A  —  1 

V  i  -  sin2  A 

V i  +  tan2  A 

Vi  +  Cot2A 

cosec  A 

• 

sin  A 

tan  A 

1 

V 1  -  cos2  A 

1 

tan  A 

VSec2  A-l 

V  i  -  sin2  A 

cos  A 

cot  A 

V  cosec2  A  —  1 

cos  A 

1 

cot  A 

1 

Vi-Sin2^l 

cot  A 

"^  cosec2  A  —  1 

sin  A 

V  l  -  cos2  A 

tan  A 

v'sec2  A  —  1 

1 

1 

sec  J. 

V  i  +  cot2  A 

cosec  A 

sec  A 

V i +  tan2  A 

Vl-sin2^ 

cos  A 

cot  A 

V  cosec2  A  —  1 

1 

1 

sec  A 

Vi+tan2^ 

cosecA 

Vi+cot2/x 

cosec  A 

sin  A 

Vi-cos2;l 

tan  A 

VSec2A-l 

15.  90°,  36°  52'  12".         16.  45c 
19.  30°,  48°  35'  25". 


17.  45°,  71°  34'.      18.  53°  V  48". 
20.  36°  52'  12",  16°  15'  36". 


Art.  9,  Page  24 

5.  A  =65°  14',  £  =  24°  46',  6  =  7.834. 

6.  303.9,  39°  47'.6,  50°  12'.4. 

7.  58°  45'  48",  a  =  1521.5,  6  =  2508.6. 

8.  ^=21°  8',  6  =  94.43,  c  =  101.24. 

9.  A  =30°  12'.2,  £  =  59°  47'.8,  c  =  116.25. 

10.  A  =  l°  21'.9,  5  =  88°  38'.1,  6=45.95. 

11.  £=41°43',a  =  241.85,  6  =  215.6. 
12    £  =  38°  41',  a  =  312.2,  c=400. 

13.  5  =  52°  39'  30",  a  =  1040.9,  6  =  1364.3. 

14.  .4  =  52°  37'.1,  5  =  37°  22'.9,  c  =  2912.1. 

15.  £  =  62°  14'  40",  a  =  1968.7,  c=4227.4. 

16.  A  =  29°  24'.9,  5  =  60°  35'.1,  6  =  43.67. 

17.  4=27°  20',  6  =  77.4,  c  =  87.1.         18.  6  =  7.4833,  A  =33°  44'.6. 
19.  c  =  8.75,  A=30°  57'.8.  20.  a  =  9.57,  6  =  11.54. 

21.  a  =  3.84,  6  =  11.37.  22.  6  =  8.9433,  ^=41°48'.6. 

83.  a  =  16.4,  c  =  22.2.  24.  c  =  14.42,  4=33°  41\4. 


130  ANSWERS 

Art.  10,  Page  25 
1.  24.  95  ft.,  12.71  ft.  2.  58.78  in. 

3.  (a)  9.239  in.     (6)  21.96  ft.     (c)  16.8  ft.     (d)  19.3  ft. 

Art.  11,  Pages  27,  28 

4.  398.2  ft.  7.  276.95  ft.  10.  86.6  ft.,  50  ft. 

5.  228  4  ft.,  258  ft.        8.  463.7  ft.  11.  749  mi. 

6.  63.9  ft.,  63.9  ft.        9.  3243.8  ft.  12.  219.45  ft. 

Art.  12,  Page  29 

2.  Base  =  187.9  ft.;  height  =  350.63  ft.;  area  =  32,943  sq.ft. 

3.  Base  =  358.21  ft.;  height  =  161.26  ft.;  area  =  28,881  sq.ft. 

4.  68°    50'    5".4,    68°    50'    5".4,    42°    19'    49,,.2;      height  =  83.93    ft.; 

area  =  2727.7  sq.ft. 

5.  56°  18'.6,  56°  18\6,  67°  22'.8;  36.06  ft.;  600  sq.ft. 

6.  105.83  ft.;  48°  35',  48°  35',  82°  50',  3175  sq.ft. 

7.  40.76  ft.,  44.9  ft.,  44.9  ft.;  area  =  815.2  sq.ft. 

8.  96.68  ft.,  79.4  ft.,  79.4  ft.;  area  =  3045.4  sq.ft. 

Art.  13,  Page  31 
1.  14.54  ft.,  16.13  ft.,  48.45  sq.ft.,  105.2  sq.ft. 

Art.  14,  Page  33 

1.  26.172,  52.345  miles;    second  ship  bears  E.  19°  42M  N.  from  first. 

2.  LB  =  14.197  miles. 

Art.  15,  Page  35 
4.  2.852  acres.  5.  12  acres,  3  roods,  6.45  poles. 

Art.  16,  Page  38 

1.  a.  For  137°:  second;  137°,  497°,  857°,  1217°. 
For  785°:  first;         65°,  425°,  785°,  1145°. 
For  3657°:  first;       57°,  417°,  777°,  1137°. 
6.  For  -240°;  second;  120°,  480°,  840°,  1200°. 
For  -337°:  first;         23°,  383°,  743°,  1103°. 
For  -7283°;  fourth;   277°,  637°,  997°,  1357°. 

3.  320°,    -60°,  130°,    -250°,  15°,  7°,  78°,  385°,  414°,    -110°,  -150°, 

200°,  257°. 

4.  410°,  30°,  220°,  -160°,  105°,  97°,  168°,  475°,  504°,  -20°,  -60° 

290°,  347°. 


ANSWERS  131 


Art.  18,  Pages  41,  42 
1.  143°  14'  22".  2.  4.03. 

3.  28°  38'  52".4,  229°  10'  59".2,  19°  5'  54".9. 

4.  90°,  60°,  45°,  30°,  120°,  540°,  -240°,  -1800°. 

.-.«.«-  /»        7T       7tf      57T     2tf     57T 

5.  1.454,  2.487.  6.  — ,  — ,  -j,  -j,  p 

7.  ^;  8°  29'.  8.  6  in. 

50 
9.  40  in.  10.  —  mins. 

71 


Art.  18a,  Page  43 
1.  7U  radians  per  sec.  2.  1 :  12 :  720. 

3.  -5-  radians  per  sec. 

o 

4.  (a)  —  radians  per  sec;  (6)  1800°  per  minute. 

6 

,    ,  ,  3960         .   ..  .        „,  132      ,. 

5.  (a)  -= —  revolutions  per  mm.;  (0)  -=—  radians  per  sec. 

In  7 

6.  39.27  ft.  per  sec. 

Art.  20,  Pages  46,  47,  48 

1.  (a)   +;  (6)   -;  (c)   +;  (d)   +;  (e)   +;  (/)   -. 

4.  (a)  First  and  second;  (6)  second  and  third; 
(c)   first  and  third;        (d)  second  and  fourth. 

5.  (a)  Third;  (b)  second;  (c)  third;  (d)  fourth. 

In  the  following  examples  the  functions  are  given  in  the  order  on 
page  40. 

A   _A   _A   _A    _A  A.  _A  A   _A   _A  A   _A 

5'        5'        4'        3'        4'   3;        5'   5'        4'        3'   4'        3* 


8. 


q    n      ,^5      _2     _V5      __2_    _^        _3_ 
».  W-    +  3    1        3>         2    '       VT'        2'  +VJ' 

_VW    __2_       VT         _2_     _3^  3 

3    '       3'  +   2    '    +Vr        2'       vT 

()        5'        5'  V    3'        4'       V        5'   5'        4'        3'  4'       3' 


132  ANSWERS 


M   -u^H    1     -u^l       -A-    A       _A_ 
W    +    5    '     5'   +    2    '   Tv^i'     2'    V2T 

\/21      2         V21  2        5  5 


5     '     5'  2    '        V2l'    2'       V2l' 

A  A  A  A  A  A    _A  A   _A   _ 3  A    _A 

5'    5'    3'    4'   3'   4  '         5'   5'        3'        4 '   3 '        4 

A  A  i  A  A  A.  _A    _A  A  A   _A   _A 

5 '    5  '    3  '   4 '    3  '   4  '         5'        5 '   3  '   4  '        3'        4* 

3        2\/J0  3  2VlO  7       _7 

10.  (a)  -?,  -      ?     ,      2ViQ,  3     ,      avl_>   3. 

V21  2^    _^H  _2_    J[  ^5_ 

()          5    '  5'          2    '  V2l'    2'  V2l' 

3              2       j5  2_       VJ3  Vl3 

(C)       VI3'  Vfs'    2'  3'  ""    2    '  ~      3    ' 


(<*) 


2  •    2'  "'     VT  "'       V3" 


V3,        *      2,   -^ 


Art.  25,  Page  62 

a2  +  62'  a2— 62 

Art.  26,  Page  66 

1.  sin  (270°-.4)  =  -cos4,  cos  (270°-, 4)  =  -sin  A, 
tan  (270° -.4)  =cqt  A,  cot  (270°  -.4)  =tan  A}' 
sec  (270°-vl)=  -csc  A,  esc  (270°-4)  -  -sec  A. 

2.  sin  (270°  +  A)=-cos4,  cos  (270°  +  A)  -sin  A, 
tan  (270°  +  4)  =  -cot4,  cot  (270°  +  A)  =  -tan  4, 
sec  (270°  +  4)=csc4,  esc  (270°  +  4)  =  -sec  A. 

Art.  27,  Page  67 

4.  (a)   -sin  73°,  -cos  17°.  (e)    -sin  10°,  -cos  80°. 
(6)  cos  28°,  sin  62°.                         (/)  cot  53°,  tan  37°. 

(c)  tan  38°  30',  cot  51°  30'.  (g)  tan  25°,  cot  65°. 

(d)  -sec  25°  10',  -esc  64°  50'.    (h)  -sin  15°,  -cos  75°. 

(0   -esc  49°  30',  -sec  40°  30'. 

5.  (a)  -.2391.  (d)  1.555.  (g)  2.458. 
(6)  -.6225.  (e)  .8391.  (h)  .6293. 
(c)    -.1007.              (/)    .7660.  (i)    -1.0724 


ANSWERS  133 

Art.  28,  Page  72 

4.  n-180°  +  (-l)n30°,  nn  +  (-l)n^;  30°,  150°,  390°,  510°. 

5.  n-  360°  ±120°,  2nsr±J-«r;  120°,  240°,  480°. 

6.  (a)  n-180°,  nn.  (g)  n-  180°-(-l)w  60°,  nx-i-l?^-. 
~.         (h)  n.360°±135°,  2mz±^ 

TV 

4 


*(fc)  n-360°±90°,  2nn±\.         (h)  n-360o±135°,  2nK±371 
(c)   n-180°,  rnz.  (i)    n- 180° +45°,  nn  +  n 


(d)  n-180°  +  60,  n^+^-.  (?)    n.l80°  +  135°,  rwr+^p 

(e)  n.l80°±90°,  mz±\.  Ck)  n.360°±30°,  2nn±~. 

A  6 

(/)  n.l80°+(-l)n45°,  nn  +  (-l)nj.         (I)   n.l80°  +  150°,  nn  +  ^. 

Art.  30,  Page  77 

3.  x  =  n7z  +  (  —  l)n—;  x = sin-1  2,  impossible. 

o 

4.  x  =  2n7r+— .  5.  y  =  2ri7t+j-. 

A  4 

6.  x=mz±^r.  7.  A=2mz±7^,  A=2mz. 

8.  st-wr,  z  =  n-360°  +  (-l)»38°10'.3;  a; ^sin-^- 1.618),  impossible. 

9.  x  =  2nn±—  -;  a;  =  cos-1  >/3^  impossible. 

o 

10.  B  =  mt±%-.  11.  x  =  n7r+^,  z=n-180o+165o47'.7. 

b  4 

12.  2/  =  n-180°  +  71°33,.8,  */  =  n-180o  +  63°  26'. 

13.  60°,  120°,  240°,  300°,  420°,  480°.  14.  x  =  30°,  x  =  150°. 

Art.  31,  Pages  80,  81 

+  V21+2V15     +V/3l5-V/2'  4    i.  I  c    2\/2  +  V/3~ 

20  '  20  '         4'  5  '  5 '  6 

Art.  32,  Page  83 

84  .   187  44  m   117  2V/2-V/3~ 

205 '   205'  125 '   125*  6 


*  In  this  instance  the  general  value  may  be  expressed;  m- 180°±90°,  mn±— . 


134  ANSWERS 

Art.  34,  Page  85 
6.   1;  y.  9.  tan-i(— i-)  ;  tan-^. 

10.  tan-1  oo,     i.e.,     (2n+ !)-£-;    cot"1 0,     i.e.,  (2n  +  l)-£-. 

Art.  35,  Pages  87,  88 

3  3  3  3  ^  3 

4.  cos2  —a: — sin2  —a;,  2  cos2  — x  —  1,  1—2  sin2  —  x;  2  sin  -^-x  cos  —a;. 


/l+cos6x  ll 

5-  V^2— ;    \- 


cos  6a; 


2 
6.  cos2  3a:— sin2  3x,  2  cos2  3a;  —  1,  1—2  sin2  3x;  2  sin  3a;  cos  Sx. 

10.  M  4;    *£,    -1;   ,$ 

Art.  36,  Pages  91,  92,  93 

5.  (1)  2  sin  4a;  cos  x.  (2)  2  cos  6A  sin  A. 

(3)  2  sin  4x  sin  2a;.  (4)  2  cos  7x  cos  2a;. 

.    mA+nB        mA—nB                    .    mx  +  ny    .     mx  —  ny 
(5)  2  sin g cos 2 '    ^   ~2sm ^sin ^-2. 

(7)  2  sin  (45° -x)  cos  (45° -4x).    (8)  2  cos  (45°  -x)  sin  (45° -3x). 

6.  (1)  J(sin8x+sin2x).  (2)  \ (sin  12a; -sin  2x). 
(3)  £(cos4z— cos  8a;).                     (4)  £(cos  14a;  +  cos  4x). 

(5)  $[cos  (m A  — nB)— cos  (mA+nB)}. 

(6)  ijcos  (nx  +  my)  +  cos  (nx— my)}. 

(7)  £(cos2x— cos  6a;).  (8)  £(cos  10a;+cos4a;). 
18.  1. 

29.  ^  =  (2n  +  l)90°,  0  =  {4rc+(-l)n}15°. 
30    a-y,  a=n-90°  +  (-l)n45°. 

31.  a  =  n.90°  +  (-l)n45°,  a-*  sin-1(-|)=n.90o  +  (-l)n(-18°  26'. 1). 

32.  0=n^r,  0=n.l8O°±9O°. 

4  ' 

33.  ^  =  {6n+(-l)»}30°,<?  =  {10n  +  (-l)Tl{18o,^  =  {10M-3(-l)nil8o. 

34.  *=-i,  *-l.  35.  a;=0,  a?«-±$.  36.  x  =  i  x=-8. 
37    ta    (A  +  B  +  C)  -  tan  ^  +tan  ^+tan  C  — tan  A  tan  B  tan  C 

1  —tan  A  tan  5  — tan  5  tan  C  —  tan  C  tan  4* 

Art.  41,  Page  100 
3.  38°  12'.8,  60°,  81°  47'.4.  4.  41°  24\7, 


ANSWERS  135 

Art.  43,  Page  103 

2.  6-70.8,  a -56.1  4.  6  =  185,  c  =  192. 

3.7.98  ft  5.  6  =  8.237,  c  =  5.464. 

Art.  44,  Page  107 

2.  5-36°  18'.4    or    143°  41'.6,    c  =  52.71    or    5.98. 

3.  Triangle  impossible. 

4.  5=46°  28',     C  =  68°32',      c  =  30.8  in. 

5.  A=48°10',     C  =  108°25',    c  =  19.1  ft.;     or 
A  =  131°  50',  C=24°  45',      c  =  8.4  ft. 

Art.  45,  Page  108 

2.  A=77°12'.9,  5=43°  30M,  c  =  29.97. 

3.  A=80°46'.4,  C  =  63°48'.6,  6  =  19.4. 

4.  #  =  33°    3'.3,  S  =  100°56'.7,  Z  =  39.6  ft. 

5.  P  =  29°41'.2,  Q  =  52°24'.4,  r  =  20in. 

Art.  46,  Page  109 

2.  P  =  19°  12',  Q  =  61°  13',  P  =  99°  35'. 

3.  48°  11'.4,  58°  24'.7,  73°  23'.9. 

4.  28°  22',  49°  43',  101°  56',  nearly. 

5.  93°  41',  67°  23',  18°  56'. 

Art.  48,  Pages  110,  111 

2.  C=66°47',  6  =  698.3,  c  =  845. 

3.  600,  421.5.  4.  Triangle  impossible. 

5.  A  =42°  53'  34",  C  =  77°  5'  54",  c  =  394.53. 

6.  (7  =  64°  24',  £  =  78°  16',  6=749.1. 

7.  B =46°  52'  10",     C  =  lll°53'  25",    c  =  1767.3;     or 
5  =  133°  7'  50",     C=25°  37'  45",       c  =  823.8. 

Art.  49,  Page  112" 

2.  £=64°  9'  3",  A  =42°  50'  57",  c  =  374. 

3.  P  =  132°  18'  27",  Q  =  U°  34'  24",  r= 67.75. 

4.  A  =  109°  15'.5,  5=45°  4'.5,  c= 440.5. 

5.  5  =  88°  2'.6,  C  =  23°  43',  a  =  8.439. 

Art.  50,  Page  113 

2.  53°  7'.8,  59°  29'.4,  67°  22'.8.  3.  A  =29°  17'  16",  5  =  31°  55'  31". 

4.  44°  48'  15",  52°  55'  56",  82°  15'  49". 

5.  #  =  25°  12'  32",  S=48°  11'  22." 


136  ANSWERS 

Art.  51,  Pages  114,  115 

3.  444.72  yd.  6.  179.28  ft. 

4.  1112.8  yd.  7.  87.88  ft. 

5.  489.29  yd.;  505.3  yd.  8.  104.08  ft. 

9.  479.3  yd.  10.  .Height  =  119.6  ft.,  AC  =  153.1  ft.;  5C-571  ft. 

11.  469.6  ft. 

Art.  52,  Page  118 

1.  (a)  536.06  sq.ft.  2.  (a)  668.7  sq.ft. 

(6)  325.7  sq.ft.  (6)   500.3  sq.ft. 

3.  Each  of  the  equal  sides  =  28.6  yd. 

4.  The  triangles  in  which  the  angles  between  the  given  sides  are  re- 

spectively 30°  17'.8  and  149°  42'.2. 

5.  154.37  yd.  on  the  road  opposite  the  right  angle. 

Art.  53,  Page  119 

1.  (a)  6§  in.,  33£  sq.in.  (c)  36  in.;  324  sq.in. 
(6)  30  in.;  360  sq.in.  (d)   90  in.;  900  sq.in. 

2.  27f  in.;  329|  sq.in.  3.  11.1  in.;  22.2  in. 
4.  Radius  =  14|  in.:  angle  =  2i§  radians  =  124°.3. 

Art.  54,  Page  121 

1.  (a)  7.67;  21.6;  13.7,  25.5,  54.8  in.  respectively. 
(6)  2.6;  6.5,  7.3,  16,  5.3  in.  respectively, 
(c)   10.48;  25.03;  22.04,  30.44,  58.11  in.  respectively. 

Art.  55,  Pages  126, 127 

4.  238,890  mi.  (approx).,  347.5.        5.  About  57'  2";  about  1:13.5. 

6.  About  93,757,000  mi. 

7.  206,265  times  the  distance  of  the  earth  from  the  sun;  3.26  yr. 

8.  9  ft.  2.6  in.  9.  76  ft.  9.5  in.  10.  4'  35". 
11.  15.708  yd.              12.  13'  1".3. 


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